ENGIN. 
LIBRARY 


UC-NRLF 


Meeh,  dept. 


PRACTICAL     CALCULATION     OF 
TRANSMISSION    LINES 


PEACTICAL  CALCULATION 

OF 

TBANSMISSION  LINES 


FOR    DISTRIBUTION    OF    DIRECT    AND    ALTERNATING 
CURRENTS  BY  MEANS  OF  OVERHEAD,  UNDER- 
GROUND, AND  INTERIOR  WIRES  FOR 
PURPOSES  OF  LIGHT,  POWER, 
AND  TRACTION 


BY 

L.  W.  ROSENTHAL,  E.E. 

o  • 

ASSOCIATE   MEMBER,  A.I.E.E. 


NEW  YORK 

MCGRAW  PUBLISHING  COMPANY 

239  WEST  39TH  STREET 
1909 


Engineering 
library 


COPYRIGHT,  1909, 

BY  THE 

McGKAW  PUBLISHING  COMPANY 
NEW  YORK 


Stanbopc  Jpresa 

F.    H.G1LSON     COMPANY 
BOSTON.     U.S.A. 


PREFACE. 


THIS  little  book  is  offered  to  the  engineering  profession  with 
the  hope  that  it  may  be  of  practical  help  in  the  rapid  and  accurate 
calculation  of  transmission  lines.  Its  existence  is  the  outcome  of 
the  belief  that  this  field  is  in  part  barren.  Its  chief  mission  is  to 
substitute  a  direct  solution  for  the  trial  method  which  was  formerly 
a  necessary  evil.  The  arrangement  of  the  formulas,  tables  and  text 
has  been  dictated  solely  by  the  needs  of  the  rapid  worker. 

All  sections  except  the  last  include  the  important  effects  of 
temperature  and  specific  conductivity,,  The  section  relating  to 
direct-current  railways  is  novel  in  the  form  of  its  tables,  and  the 
methods  outlined  in  it  have  been  found  rapid  and  comprehensive., 
The  alternating-current  division  presents  a  new  and  original  method 
for  the  solution  of  these  problems.  It  is  the  only  method  known  to 
the  author  which  determines  the  size  of  wire  directly  from  the  volt 
loss  in  the  line,  and  it  also  possesses  unique  features  of  scope,  accu- 
racy and  simplicity.  The  chapter  on  single-phase  railways  is  in 
accord  with  most  of  the  consistent  data  published  on  the  subject, 
although  further  accurate  investigations  of  installed  lines  may 
modify  to  some  extent  the  present  accepted  values. 

The  author  desires  to  call  particular  attention  to  the  fallacies  of 
some  familiar  methods  of  calculating  alternating-current  transmis- 
sion lines  which  heretofore  have  been  in  common  use.  It  will  be 
evident  from  Tables  11,  28,  and  36  that  their  results  are  wholly 
erroneous  under  certain  practical  conditions,  and  indicate  wires 
which  may  be  entirely  at  variance  with  the  specified  requirements. 

The  scope  of  this  book  has  been  confined  to  methods  of  calcula- 
tion. Hence,  the  most  desirable  limits  of  line  losses  are  not 
discussed,  but  the  tables  are  sufficiently  extended  to  cover  all  cases 
likely  to  arise  in  practice.  There  is  no  discussion  of  the  character- 
istics of  alternating-current  transmission  lines,  although  the  tables 
of  wire  factors  render  apparent  many  of  their  important  features. 
Furthermore,  the  book  does  not  include  either  the  determination  of 

iii 


254556 


iv  PREFACE 

the  size  of  conductors  for   conditions  of  maximum  economy  or  the 
consideration  of  alternating-current  circuits  in  series. 

The  preparation  of  the  tables  has  involved  thousands  of  calcula- 
tions, but  thorough  checks  by  the  methods  of  differences  and  curve 
plotting  have  probably  eliminated  almost  all  errors  of  material 
influence.  However,  some  discrepancies  may  have  crept  in,  and  the 
author  would  be  glad  to  learn  of  them. 

L.  W.  R. 

NEW  YORK  CITY.  December,  1908. 


CONTENTS. 


CHAPTER  I.— DIRECT-CURRENT  DISTRIBUTION  FOR  LIGHT 

AND  POWER. 
PAR.  PAGES 

1.  INTRODUCTION 6^ 

2.  PROPERTIES  OF  CONDUCTORS : 5 

3.  CURRENT-CARRYING  CAPACITY 6 

4.  PARALLEL  RESISTANCE  OF  WIRES 9 

5.  GIVEN  ITEMS 10 

6.  FORMULAS 10 

7.  AMPERE-FEET 10 

8.  EXAMPLES.  . 11 


CHAPTER   II.  —  DISTRIBUTION  FOR  DIRECT-CURRENT 
RAILWAYS. 

9.  INTRODUCTION 16 

10.  RESISTANCE  OF  RAILS 16 

11.  PARALLEL  RESISTANCE  OF  RAILS  AND  FEEDERS 17 

12.  NEGATIVE  CONDUCTORS 18 

13.  POSITIVE  CONDUCTORS 18 

14.  RESISTANCE  OF  CIRCUIT 19 

15.  GIVEN  ITEMS 19 

16.  EXAMPLES  .  .  20 


CHAPTER  III.  —  ALTERNATING-CURRENT  TRANSMISSION  BY 
OVERHEAD  WIRES. 

17.  INTRODUCTION 29 

18.  OUTLINE  OF  METHOD 30 

19.  RANGE  OF  APPLICATION 31 

20.  MAXIMUM  ERROR 31 

21.  TRANSMISSION  SYSTEMS 32 

22.  BALANCED  LOADS 33 

23.  TEMPERATURE 33 

24.  SPECIFIC  CONDUCTIVITY 33 

25.  SOLID  AND  STRANDED  CONDUCTORS 33 

26.  SKIN  EFFECT 33 

27.  WIRE  SPACING 34 

v 


vi  CONTENTS 

PAR.  PAGE 

28.  ARRANGEMENT  OF  WIRES 34 

29.  FREQUENCY 34 

30.  MULTIPLE  CIRCUITS 34 

31.  CURRENT-CARRYING  CAPACITY 34 

32.  TRANSMISSION  VOLTAGE 35 

33.  VOLT  LOSS 35 

34.  POWER  TRANSMITTED 35 

35.  POWER  LOSS 36 

36.  POWER-FACTOR 36 

37.  WIRE  FACTOR 36 

38.  GIVEN  ITEMS 37 

39.  SIZE  OF  WIRE 37 

40.  PER  CENT  VOLT  LOSS 37 

41.  CHARGING  CURRENT 38 

42.  CAPACITY  EFFECTS 38 

43.  EXAMPLES  .  .  39 


CHAPTER  IV.  —  ALTERNATING-CURRENT  TRANSMISSION  BY 
UNDERGROUND   CABLES. 

44.  INTRODUCTION 63 

45.  MAXIMUM  ERROR 63 

46.  TEMPERATURE 64 

47.  PROPERTIES  OF  CONDUCTORS 64 

48.  THICKNESS  OF  INSULATION 64 

49.  CURRENT-CARRYING  CAPACITY 64 

50.  CAPACITY  EFFECTS 65 

51 .  EXAMPLES 65 

CHAPTER   V.  —  INTERIOR    WIRES   FOR    ALTERNATING-CURRENT 

DISTRIBUTION. 

52.  INTRODUCTION 75 

53.  PROPERTIES  OF  CONDUCTORS 75 

54.  SPACING  OF  WIRES 75 

55.  AMPERE-FEET 76 

56.  EXAMPLES 76 

CHAPTER  VI.  —  DISTRIBUTION  FOR  SINGLE  PHASE 
RAILWAYS. 

57.  INTRODUCTION 85 

58.  METHOD  OF  CALCULATION 85 

59.  IMPEDANCE  OF  RAIL 85 

60.  PERMEABILITY  OF  RAIL 86 

61.  IMPEDANCE  AND  WEIGHT  OF  RAIL 86 

62.  IMPEDANCE  OF  RAIL  AND  FREQUENCY 86 


CONTENTS  vii 

PAR.  PAGE 

63.  FORMULA  FOR  RAIL  IMPEDANCE 86 

64.  POWER-FACTOR  OF  TRACK, , 87 

65.  HEIGHT  OF  TROLLEY 87 

66.  EFFECT  OF  CATENARY  CONSTRUCTION  ,  . . 87 

67.  IMPEDANCE  OF  COMPLETE  CIRCUIT 87 

68.  MULTIPLE  TRACKS  ..  =  .,.. , 88 

69.  EXAMPLES  .  .                           ,.....-..* _ 88 


TABLES. 


CHAPTER   I.  — DIRECT-CURRENT    DISTRIBUTION    FOR  LIGHT 

AND  POWER. 
No.  PAGE 

1.  Values  of  T  for  copper  and  aluminum 6 

2.  Values  of  F  for  copper  and  aluminum . 6 

3.  Properties  of  copper  and  aluminum , 7 

4.  Ampere-feet  per  volt  drop  and  current-carrying  capacity 8 

5.  Values  of  H  for  copper  or  aluminum. , 8 

Formulas  for  direct-current  wiring 14 

6.  Values  of  a  for  copper  and  aluminum. 15 

CHAPTER  II.— DISTRIBUTION  FOR  DIRECT  CURRENT 
RAILWAYS. 

7.  Values  of  Tl  for  steel 17 

8.  Equivalents  of  copper  of  100  per  cent  conductivity. .  , 20 

9.  Resistance  to  direct  current  of  one  steel  rail 24 

Formulas  for  direct-current  railway  circuits , 25 

10.  Values  of  A  for  wires  and  rails „ 26 

CHAPTER  III.  — ALTERNATING-CURRENT  TRANSMISSION  BY 
OVERHEAD  WIRES. 

11.  Error  in  per  cent  of  volt  loss , 31 

12.  Maximum  error  in  per  cent  of  true  values  at. 20°  cent 32 

13.  Values  of  c  for  overhead  wires 39 

14.  Reactance  factors ,...., 39 

Formulas  for  a.  c.  transmission  by  overhead  wires 50 

15.  Values  of  volt  loss  factors . .  51 

16.  Values  of  A  for  balanced  loads 52 

17.  Values  of  B  for  balanced  loads. , 52 

18.  Values  of  M  for  overhead  copper  wires  at  15  cycles  per  second 53 

19.  Values  of  M  for  overhead  copper  wires  at  25  cycles  per  second 54 

20.  Values  of  M  for  overhead  copper  wires  at  40  cycles  per  second 55 

21.  Values  of  M  for  overhead  copper  wires  at  60  cycles  per  second. .....  56 

22.  Values  of  M  for  overhead  copper  wires  at  125  cycles  per  second 57 

23.  Values  of  M  for  overhead  aluminum  wires  at  15  cycles  per  second. . .  58 

24.  Values  of  M  for  overhead  aluminum  wires  at  25  cycles  per  second. . .  59 

25.  Values  of  M  for  overhead  aluminum  wires  at  40  cycles  per  second. . .  60 

26.  Values  of  M  for  overhead  aluminum  wires  at  60  cycles  per  second. . .  61 

27.  Values  of  M  for  overhead  aluminum  wires  at  125  cycles  per  second. .  62 

ix 


X  TABLES 

CHAPTER  IV.  —  ALTERNATING-CURRENT  TRANSMISSION  BY 
UNDERGROUND  CABLES. 

No.  PAGE 

28.  Error  in  per  cent  of  true  volt  loss ' 63 

29.  Maximum  error  in  per  cent  of  true  values  at  20°  cent 64 

30.  Values  of  c  for  underground  cables 65 

Formulas  for  a.  c.  transmission  by  underground  cables 71 

31.  Values  of  VQ'"  =  F0  (1  -  0.01  F0) 72 

32.  Values  of  A  for  balanced  loads 72 

33.  Values  of  B  for  balanced  loads 72 

34.  Values  of  M  for  multiple  conductor  copper  cables 73 

35.  Values  of  M  for  multiple  conductor  copper  cables 74 

CHAPTER  V.  —  INTERIOR   WIRES  FOR    ALTERNATING-CURRENT 
DISTRIBUTION. 

36.  Error  in  per  cent  of  true  volt  loss 76 

Formulas  for  a.  c.  interior  wiring 81 

37.  Values  of  a  and  b  for  balanced  loads 82 

38.  Values  of  B  for  balanced  loads 82 

39.  Values  of  M  for  copper  wires  in  interior  conduits 83 

40.  Values  of  M  for  copper  wires  in  molding  or  on  cleats «...  84 

CHAPTER  VI.— DISTRIBUTION    FOR    SINGLE-PHASE   RAILWAYS. 

41.  Test  and  calculated  values  of  impedance  per  mile 88 

Formulas  for  single-phase  railway  circuits 92 

42.  Values  of  M  for  single-phase  railway  circuits 93 


PART  I. 

DIRECT-CURRENT   DISTRIBUTION 

BY   MEANS   OF 
OVERHEAD,  UNDERGROUND   AND   INTERIOR   WIRES 

FOR   PURPOSES   OF 
LIGHT,  POWER  AND  TRACTION 


DIRECT-CURRENT   DISTRIBUTION 


CHAPTER  I. 
DIRECT-CURRENT  DISTRIBUTION  FOR  LIGHT  AND  POWER. 

1.  Introduction.  —  Problems    in   direct-current    transmission 
and  distribution  are  relatively  simple.     The  same  formula  covers 
all  conditions  of  installation  and  operation  whether  by  overhead, 
underground  or   interior    wires.     The    formulas    give    accurate 
results,  and  all  items  of  influence  are  easily  included.     The  tables 
are  concise  but  comprehensive   and  will  cover  almost   all  the 
usual  and  unusual  requirements  of  varied  practice. 

2.  Properties  of  Conductors.  —  The  resistance  of  stranded  and 
solid  conductors  of  the  same  cross  section  and  length  is  practically 
the  same.     Table  3,  page  8,  gives  the  properties  of  wires  at  20° 
cent,  or  68°  fahr.  for  copper  of  100  per  cent  and  aluminum  of 
62  per  cent  conductivity  in  Matthiessen's  standard  scale.     The 
resistance  at  any  other  temperature  and  conductivity  may  be 
found  for  copper  or  aluminum  from  formulas  (1)  and  (2). 


v 

Ohms  resistance  per  1000  feet    =      >        * 


.     . 

S 

Ohms  resistance  per  mile  =  54/700  X  T  ^     ^ 

o 

S  =  Cross  section  of  metal  in  circular  mils. 
T  =  Temperature  factor.     Table  1,  page  6. 

Thus,  at  40°  cent,  the  resistance  per  mile  of  No.  1  copper  wire 
of  98  per  cent  conductivity  is 


6 


TRANSMISSION  CALCULATIONS 
Table  i. —  Values  of  T  for  Copper  and  Aluminum. 


Temperature 

Copper. 

Alumi- 
num 

in 

degrees. 

Conductivity  in  Matthiessen's  standard  scale. 

Cent. 

Fahr. 

1.00 

0.99 

0.98 

0.97 

0.96 

0.62 

0 

32 

0.926 

0.936 

0.946 

0.955 

0.965 

.495 

10 

50 

0.963 

0.973 

0.983 

0.993 

1.003 

.553 

20 

68 

1.000 

1.010 

1.020 

1.031 

1.042 

.613 

30 

86 

1.038 

1.048 

1.059 

1.069 

1.081 

.674 

40 

104 

1.077 

1.088 

1.099 

1.110 

1.122 

.738 

50 

122 

1.117 

1.128 

1.140 

1.150 

1.163 

.802 

| 

For  copper  or  aluminum  ot  other  conductivity   divide  value  under  1.00  by  given  con- 
ductivity. 

The  weight  of  a  unit  length  of  copper  or  aluminum  wire  may  be 
calculated  from  the  following: 

Pounds  per  unit  length  =  FS (3) 

F  =  Weight  factor.     Table  2,  page  6. 

S  =  Cross  section  of  metal  in  circular  mils. 

Table  2. —  Values  of  JFfor  Copper  and  Aluminum. 


Unit  Length. 

Copper  Wires. 

Aluminum  Wires 

Solid. 

Stranded. 

Solid. 

Stranded. 

1000  feet  
Mile  

0.00303 
0.0160 

0.00305 
0.0161 

0.00091 
0.00482 

0.00092 
0.00486 

3.  Current-carrying  Capacity.  —  Table  4,  page  8,  gives  the 
current-carrying  capacity  of  wires  under  various  conditions. 
Overhead  wires  larger  than  No.  0  have  been  considered  stranded, 
while  those  smaller  than  No.  0  have  been  taken  solid.  The  values 
for  interior  wires  have  been  taken  from  the  National  Electric 
Code.  The  results  for  bare  and  insulated  overhead  wires  are 
calculated  from  formula  (4)  for  copper  of  100  per  cent  conduc- 
tivity and  a  temperature  rise  of  40°  cent,  or  72°  fahr.  The 


DIRECT-CURRENT    DISTRIBUTION 


Table  3. —  Properties  of  Copper  and  Aluminum  at  20°  Cent,  or 
68°  Fahr.  Conductivity  in  Matthiessen's  Standard  Scale ; 
Copper  100  Per  Cent,  Aluminum  62  Per  Cent. 


Copper. 

Aluminum. 

B.&S. 

Area, 

Diame- 

f  cr 

Ohms 

Pounds 

Ohms 

Pounds 

A.W.G. 

Cir.  Mils. 

tcr, 
Inches. 

Resistance. 

Weight. 

Resistance. 

Weight. 

1000  Ft. 

Mile. 

1000  Ft. 

Mile. 

1000  Ft. 

Mile. 

1000  Ft. 

Mile. 

Bare  Stranded  Wires. 

1,000,000 

1.152 

0.0104 

0.0547 

3050 

16100 

0.0167 

0.0882 

920 

4860 

800,000 

1.035 

0.0129 

0.0683 

2440 

12880 

0.0209 

0.110 

736 

3890 

700,000 

0.963 

0.0148 

0.0781 

2140 

11270 

0.0239 

0.126 

644 

3400 

600,000 

0.891 

0.0173 

0.0911 

1830 

9660 

0.0278 

0.147 

552 

2920 

500,000 

0.819 

0.0207 

0.109 

1530 

8050 

0.0334 

0.176 

460 

2430 

400,000 

0.728 

0.0259 

0.137 

1220 

6440 

0.0417 

0.220 

368 

1940 

300,000 

0.630 

0.0345 

0.182 

915 

4830 

0.0557 

0.294 

276 

1460 

250,000 

0.590 

0.0414 

0.219 

762 

4030 

0.0668 

0.353 

230 

1220 

0000 

211,600 

0.530 

0.0489 

0.258 

645 

3410 

0.0790 

0.417 

195 

1030 

000 

167,800 

0.470 

0.0617 

0.326 

513 

2710 

0.0997 

0.526 

154 

816 

00 

133,100 

0.420 

0.0778 

0.411 

406 

2140 

0.126 

0.664 

122 

647 

0 

105,500 

0.375 

0.0981 

0.518 

322 

1700 

0.158 

0.837 

97.1 

513 

1 

83,690 

0.330 

0.124 

0.653 

255 

1350 

0.200 

1.06 

77.0 

407 

2 

66,370 

0.291 

0.156 

0.824 

203 

1070 

0.252 

1.33 

61.0 

323 

3 

52,630 

0.261 

0.197 

1.04 

160 

845 

0.318 

1.68 

48.5 

256 

4 

41,740 

0.231 

0.248 

1.31 

127 

671 

0.401 

2.12 

38.5 

203 

Bare  Solid  Wires. 

0000 

211,600 

0.460 

0.0489 

0.258 

641 

3380 

0.0790 

0.417 

193 

1020 

000 

167,800 

0.410 

0.0617 

0.326 

508 

2680 

0.0997 

0.526 

153 

809 

00 

133,100 

0.365 

0.0778 

0.411 

403 

2130 

0.126 

0.664 

121 

640 

0 

105,500 

0.325 

0.0981 

0.518 

320 

1690 

0.158 

0.837 

96.1 

508 

1 

83,690 

0.289 

0.124 

0.653 

253 

1340 

0.200 

1.06 

76.3 

403 

2 

66,370 

0.258 

0.156 

0.824 

201 

1060 

0.252 

1.33 

60.4 

320 

3 

52,630 

0.229 

0.197 

1.04 

159 

841 

0.318 

1.68 

48.0 

253 

4 

41,740 

0.204 

0.248 

1.31 

126 

667 

0.401 

2.12 

38.1 

201 

5 

33,100 

0.182 

0.313 

1.65 

100 

529 

0.505 

2.67 

30.2 

159 

6 

26,250 

0.162 

0.394 

2.08 

79.5 

420 

0.637 

3.37 

23.9 

127 

8 

16,510 

0.129 

0.627 

3.31 

50.0 

264 

1.01 

5.35 

15.1 

79.5 

10 

10,380 

0.102 

0.997 

5.27 

31.4 

166 

1.61 

8.51 

9.5 

50.0 

12 

6,530 

0.0808 

1.59 

8.37 

19.8 

104 

2.56 

13.5 

6.0 

31.8 

14 

4,107 

0.0641 

2.52 

13.3 

12.4 

65.6 

4.07 

21.5 

3.7 

19.8 

16 

2,583 

0.0508 

4.01 

21.2 

7.8 

41.3 

6.48 

34.2 

2.4 

12.4 

18 

1,624 

0.0403 

6.37 

33.7 

4.9 

26.0 

10.3 

54.4 

1.5 

7.8 

8 


TRANSMISSION  CALCULATIONS 


Table  4.  —  Ampere-Feet  per  Volt   Drop  and   Current-Carrying 

Capacity. 


Ampere-Feet 

Current-Carrying  Capacity  in  Amperes  per  Wire. 

Cir.Mils 
or 
A.W.G. 

per  Volt  Drop 
at  20° 

Cent* 

Interior  Wires. 
Nat'l   Elec. 
Code.t 

Overhead 
Bare  Wires. 

Overhead 
Wires,  Rubber 
Insulation. 

Underground 
Copper  Cables. 

Copper. 

Alumi- 
num. 

Rub- 
ber. 

W'th'r- 
proof. 

Copper. 

Alumi- 
num. 

Copper. 

Alumi- 
num. 

1-Condr 

3-Condr 

1000000 

48,100 

29,900 

650 

1000 

2460 

1950 

1090 

870 

690 

800000 

38,700 

23,900 

550 

840 

2080 

1670 

930 

730 

610 

700000 

33,800 

20,900 

500 

760 

1880 

1500 

840 

660 

560 

600000 

28,900 

18,000 

450 

680 

1670 

1330 

750 

590 

500 

500000 

24,100 

15,000 

390 

590 

1480 

1170 

660 

520 

450 

400000 

19,300 

12,000 

330 

500 

1240 

980 

550 

440 

390 

300000 

14,500 

9,000 

270 

400 

990 

790 

440 

350 

320 

250 

250000 

12,100 

7490 

240 

350 

900 

720 

400 

320 

290 

220 

0000 

10,200 

6330 

210 

312 

760 

610 

340 

270 

260 

190 

000 

8110 

5020 

177 

262 

630 

500 

280 

230 

220 

170 

00 

6430 

3970 

150 

220 

540 

430 

240 

190 

190 

150 

0 

5100 

3170 

127 

185 

460 

370 

200 

160 

170 

130 

4030 

2500 

107 

156 

350 

280 

150 

120 

150 

110 

3210 

1980 

90 

130 

300 

250 

120 

100 

120 

94 

2540 

1570 

76 

110 

250 

210 

100 

84 

110 

81 

2010 

1250 

65 

92 

210 

170 

88 

69 

91 

68 

1600 

990 

54 

77 

170 

140 

74 

59 

76 

57 

1270 

785 

46 

65 

150 

120 

63 

50 

64 

48 

8 

798 

495 

33 

46 

100 

82 

44 

35 

45 

10 

502 

311 

24 

32 

12 

314 

195 

17 

23 

14 

198 

123 

12 

16 

16 

125 

77 

6 

8 

18 

78 

49 

3 

5 

*  Product  of  feet  of  one  wire  and  amperes  for  total  drop  of  one  volt  in  both  wires, 
t  These  values  are  for  copper  wires;  for  aluminum  wires,  multiply  by  0.84. 

Table  5.  —  Values  of  H  for  Copper  or  Aluminum. 


Type  of  Installation. 

Solid 
Wires. 

Stranded 
Wires. 

389 

342 

205 

190 

Overhead  wires   rubber  insulation  

165 

152 

DIRECT-CURRENT    DISTRIBUTION  9 

current  per  wire  for  lead-covered  underground  cables  is  based  on 
tests  for  a  temperature  rise  from  70°  fahr.  initial  to  150°  fahr. 
final.*  Formula  (4)  may  be  solved  for  C  to  find  the  temperature 
rise  under  given  conditions. 


Amperes  per  wire  =  H  \~r  •   .....  (4)f 

d  =  Diameter  of  wire  in  inches. 

C  =  Temperature  rise  in  degrees  centigrade. 

H=  Heat  factor.     Table  5,  page  8. 

T  =  Temperature  factor  at  final  temperature.     Table  1,  page  6. 

The  size  of  wires  is  sometimes  determined  by  their  current- 
carrying  capacity,  especially  in  interior  work  where  the  runs  are 
short.  For  longer  lines  it  is  usually  advisable  to  calculate  the 
loss  and  then  note  that  the  wire  has  sufficient  carrying  capacity, 
while  for  shorter  stretches  it  is  often  better  to  select  a  wire  of 
proper  current  capacity  and  then  find  by  calculation  whether 
the  loss  is  within  the  specified  limit. 

4.  Parallel  Resistance  of  Wires.  —  The  parallel  resistance  at 
any  temperature  for  a  number  of  wires  of  any  conductivity  is 
given  by  the  following  formulas: 

i  n 

Ohms  resistance  per  1000  feet  =  -  u> 


+    1  + 
T         T 
12 


£* 

Ohms  resistance  per  mile          =  -  "/ 


(5) 


.  .  . 

T         T 

1  i        •*•  2 

Where  all  the  wires  have  the  same  temperature  and  conductivity, 
formulas  (5)  reduce  to  those  below. 

Ohms  resistance  per  1000  feet    -     10'350  X  T    . 

S,  +  S2  +  •  •  - 

Ohms  resistance  per  mile  =     54r700  X  T     ^ 

Sl  +  S2  4-  •  •  • 
In  formulas  (5)  and  (6) 
Sit  S2  =  Circular  mils  in  respective  wires. 
T,  Tl}  T2=  Temperature  factors  of  respective  wires.   Table  1, 
page  6. 

*  Standard  Underground  Cable  Co.'s  Handbook  No.  XVII,  page  192. 
t  Based  on  formulas  in  Foster's  "  Electrical  Engineers'  Pocketbook,"  1908, 
page  208. 


10  TRANSMISSION  CALCULATIONS 

Thus,  the  resistance  per  mile  at  30°  cent,  of  one  No.  00  trolley 
wire  of  97  per  cent  conductivity  in  parallel  with  one  1,000,000- 
cir.  mil  aluminum  feeder  of  62  per  cent  conductivity  is 
54,700 


133,100       1,000,000 
1.069  1.674 


=  0.0758  ohm. 


Similarly,  the  parallel  resistance  per  1000  feet   at  40°  cent,  of 
one  No.  0  and  one  No.  4  copper  wires  of  97  per  cent  conductivity 

18  10,350  X  1.110    =  Q  078Q  ohm 

105,500  +  41,740 

5.  Given  Items.  —  The  problem  may  require  the  determina- 
tion of  the  size  of  wire  from  the  volt  loss,  or  the  volt  loss  from 
the  given  size  of  wire.     In  either  case  the  other  required  items 
are  conductor  metal  and  temperature,  current  and  distance  of 
transmission.     If  power  in  watts  is  specified,  then  the  voltage 
at  load  or  source  must  also  be  given. 

The  term  "  source  "  is  used  in  this  book  to  designate  the  point 
from  which  the  circuit,  or  the  part  of  circuit  under  consideration, 
starts.  Thus  in  direct-current  distribution  it  may  signify  the 
generator,  rotary  converter,  storage  battery,  connection  to  main, 
to  feeder  or  to  sub-feeder,  or  simply  a  certain  point  of  the  circuit. 

6.  Formulas.  —  Formulas  for  the  complete  solution  of  direct- 
current  problems  are  given  on  page  14,  and  the  method  of  pro- 
cedure will  be  clear  from  the  arrangement  of  the  table.     The 
size  of  wire  may  be  determined  from  the  volt  drop,  or  from  the 
per  cent  volt  drop,  which  in  direct-current  systems  is  always  equal 
to  the  per  cent  power  loss.     The  value  of  a  is  found  in  Table  6, 
page  15,  and  the  size  of  each  wire  is  noted  from  Table  3,  page  7, 
corresponding  to  the  required  circular  mils. 

The  per  cent  volt  drop  is  expressed  in  terms  of  power  as  well  as 
current,  so  that  problems  involving  voltage  at  source  and  watts 
at  load  may  be  solved  without  preliminary  approximation.  It 
should  be  carefully  noted  that  percent  loss,  V  or  V0,  is  expressed  as 
a  whole  number,  and  that  the  length  of  transmission,  I,  is  the  dis- 
tance from  source  to  load,  which  is  the  same  as  the  length  of  one 
wire.  In  formula  (14),  r  is  the  resistance  per  foot  of  one  wire, 
equal  to  the  values  in  Table  3,  columns  4  or  8,  divided  by  1000. 

7.  Ampere-Feet.  —  The  size  of  wires  may  be  readily  deter- 
mined from  the  ampere-feet  per  volt  drop  as  given  by  formula  (7). 


DIRECT-CURRENT    DISTRIBUTION  11 

The  wire    having  a  corresponding   value   is   then   noted   from 
Table  4,  page  8. 

Ampere-feet  per  volt  drop  =  — (7) 

/  =  Amperes. 

I   =  Distance  from  source  to  load,  in  feet. 

v  =  Drop  in  volts. 

The  values  in  Table  4  are  for  copper  and  aluminum  of  100 
per  cent  and  62  per  cent  conductivity,  respectively,  at  20°  cent. 
For  other  conductivities  or  temperatures,  formula  (8),  page  14, 
is  more  convenient. 

In  case  of  distributed  loads,  as  in  interior  lighting  work,  II  is 
the  sum  of  the  products  given  by  each  load,  /,  multiplied  by 
its  respective  distance,  I.  Thus  if  6  amperes  are  to  be  trans- 
mitted 25  ft.,  3  amperes  50  ft.,  and  2  amperes  100  ft., 

II  =  6  X  25  +  3  X  50  +  2  X  100  =  500  ampere-feet. 

8.  Examples.  —  Examples  in  practice  may  take  innumerable 
forms,  but  the  method  of  procedure  in  any  case  will  be  clear 
from  the  table  of  formulas  on  page  14.  The  following  problems 
are  typical  and  serve  to  illustrate  the  simplicity  of  the  calcu- 
lations for  direct-current  distribution.  . 

Example  i.  —  A  copper  circuit  of  97  per  cent  conductivity  is 
to  deliver  200  amperes  for  a  distance  of  1000  feet  with  a  loss  of 
10  volts  at  40°  cent. 

GIVEN  ITEMS. 

I  =  200  amperes;         v  =  10  volts;         I  =  1000  feet. 
From  Table  6,  page  15,  a  =  23.0. 

REQUIRED  ITEMS. 

Size  of  each  wire.  —  From  (8), 

«  =  23  X  200  X  1000  =  46Q  000  ci] 
10 

Use  500,000  cir.  mil  wires. 


Volt  drop.  —  From  (9), 

_  23  X  200  X  1000 
500,000 


=  9.2  volts. 


12  TRANSMISSION   CALCULATIONS 

Watt  loss.  —  From  Tables  1  and  3, 

T  =  1.110,  and  r  =  0.0000207  ohm. 
From  (14), 
p  =  2  X  1.11  X  0.0000207  X  (200)2  X  1000  =  1840  watts. 

By  assuming  the  voltage  at  the  load  at  any  value,  say  100, 
the  per  cent  drop  and  per  cent  power  loss  are  found  to  be  the 
same,  9.2  per  cent. 

Example  2.  —  A  motor  is  to  take  25  kw.  at  a  distance  of  240 
feet  with  a  loss  of  5  per  cent  of  the  110  volts  generated,  while  the 
circuit  is  to  consist  of  copper  wires  of  98  per  cent  conductivity 
with  a  temperature  of  30°  cent. 

GIVEN  ITEMS. 

w=  25,000  watts;  e0=  110  volts;  70=5per  cent;  Z  =  240  feet. 
From  (19),  v=  0.05  X  110=  5.5  volts. 


From  Table  6,  page  15,       a  =  21.9. 

REQUIRED    ITEMS. 

Size  of  each  wire.  —  From  (8), 

s  =  21.9  X  239  X  240  = 
5.5 

Use  No.  0000  wires,  for  which  S=  211,600  cir.  mils. 
Per  cent  volt  drop.  —  From  (11), 

V  "'  =      2L9  X  25,000  X  240       =  .  - 
0.01  X  211,600  X  (HO)2 

From  Table  31,  page  72,      70=  5.4  per  cent. 

Volt  drop.  —  From  (19),         v  =  0.054  X  110  =  5.9  volts. 

Volts  at  load.—  From  (13),   e=  110  -  5.9  =  104.1  volts. 

Wattloss.—  From  (14),  p  =  °-054  X  25'QOQ  =  1430  watts. 

1  ~~  U.Uo4 

Watts  at  generator.  —  From  (15), 

w0=  25,000  +  1430=  26,430  watts. 

From  Table  4,  page  8,  it  is  seen  that  the  wire  should  have 
weatherproof  insulation  or  else  be  increased  to  250,000  cir,  mils. 


DIRECT-CURRENT    DISTRIBUTION  13 

Example  3.  —  The  load  on  a  feeder  is  to  consist  of  20  amperes 
at  50  feet,  25  amperes  at  100  feet,  and  40  amperes  at  150  feet, 
from  the  source.  Calculate  the  required  size  of  a  uniform  circuit 
of  100  per  cent  conductivity  for  a  total  loss  of  2  volts  at  20°  cent. 

From  (7),   Ampere-feet  per  volt  drop 

20  X  50  +  25  X  100  X  40  X  150 


2 

From  Table  4,  page  8,  the  required  size  of  each  wire  is  No.  0 
for  copper  and  No.  000  for  aluminum. 

Example  4.  —  Copper  mains  of  98  per  cent  conductivity  are  to 
deliver  500  amperes  to  a  point  550  feet  from  a  rotary  converter 
with  a  loss  of  3  per  cent  of  the  voltage  at  load.  Calculate  the 
size  of  wire  of  98  per  cent  conductivity  and  at  50°  cent,  if  220 
volts  are  generated. 

GIVEN  ITEMS. 

7  =  500  amperes;  e0=220  volts;  V  =  3  per  cent;  Z  =  550  feet. 


From  (18),  v  =  0^03  X          -  6.4  volts. 

From  Table  6,  page  15,  a  =  23.6. 

REQUIRED  ITEMS. 

Size  of  each  wire.  —  From   (8), 

a  =  23.6  X  500  X  550  _  1;010>000  cir 

Use  1,000,000  cir.  mils. 


Volt  drop.  -  From  (9),  t;  =  23'6  X^  °*  55°  =  6.5  volts. 

1,UUU,UUU 

Per  cent  volt  drop.  —  From   (19),  70  =  -5A  =  2.95  per  cent. 

2^  ,2\j 

Volts  at  load.  —  From  (13),     e  =  220  -  6:5=  213.5  volts. 

Watts  at  load.  —  From  (16),  w  =  213.5  X  500  =  106,750  watts. 

Example  5.  —  Find  the  combined  resistance  of  1500  feet  of 
500,000  cir.  mils  of  98  per  cent  copper  wire  at  20°  cent,  in  parallel 
with  the  same  length  of  1,000,000  cir.  mils  of  aluminum  wire  of 
62  per  cent  conductivity  at  30°  cent. 

From  (5)  and  Table  1, 


1.02  1.674 


TRANSMISSION  CALCULATIONS 


Formulas  For  Direct-Current  Wiring. 


Required  Items. 

For  Voltage  Given  at  Load. 

For  Voltage  Given  at  Source. 

Circular  mils  in  each 
wire. 

a  from  Table  6,  page  15. 

0- 

^  .                               .    (8) 

V 

Then  find  wire  from  Table  3,  page  7. 

Volt  drop. 

V  = 

an                         9 

s 

Per  cent  volt  drop 
or 
Per  cent  power  loss. 

V         a11             v       (10} 

Yf"       awl          (in 

1  °        0.01  Se*  • 

0.01  Se      0.01  e'v 

F0  from  Table  31,  page  72. 

Volts  at  source  or  load. 

e0=e  +  v  =  e(l  +  0.01  F) 

(12) 

e=e0-u=e0(l-0.01F0) 
(13) 

Watt  loss. 

T  from  Table  1,  page  6.      r  from  Table  3,  page  7. 

«       °  Trin       OOlFw)          0.01  Fnw                      ,-.., 

l-0.01Fo  ' 

Watts  at  source. 

MJO  =  W  +  p  =  IV  (1   +  0. 

f)1F^                 w                tlfo 

l-0.01Fo  * 

Amperes. 

I       w           (16) 

7                w                  (17) 

j.               .    .    ^iw; 

e0(l  -0.01F0)  ' 

When  given  V,  find  v  from  v  =  0.01  Ve  =  0.01  Fe0  /(I  +  0.01  F).    .    .   (18) 
When  given  F0,  find  v  from  v  =  0.01  F0e0  =  0.01  F0e/  (1  -  0.01  F0)  .    .    (19) 
When  given  to,  find  I  from  (16)  or  (17)  above. 

NOTATION. 


a  =  Resistance  factor.    Table  6,  page  15. 
e  =  Volts  at  load. 
e0  =  Volts  at  source. 
/  =  Amperes  from  source. 
I  =  Distance    from   source  to    load,    in 

feet. 

p  =  Total  power  loss  in  watts, 
r  =  Resistance  per   foot   of   one  wire, 

in  ohms.    Tablet,  page  7. 
S  =  Circular  mils  in  each  wire. 
T  =  Temperature     factor.       Table     1, 

page  6. 


xVolt  drop  in  per  cent  of  volts  at 
v=\     load. 

f  Power  loss  in  per  cent  of  power  at 
V     load. 

(Volt  drop  in  per  cent  of  volts   at 
source. 
~  f  Power  loss  in  per  cent  of  power  at 

^     source. 

O"'  =  Volt  loss  factor.    Table  31,  page  72. 
v  —  Volt  drop. 
w  =  Watts  at  load. 
«?0  =  Watts  at  source. 


DIRECT-CURRENT  DISTRIBUTION 


15 


Table  6.  —  Values  of  a  For  Copper  and  Aluminum. 


Temperature 
in 
degrees. 

Copper. 

Alumi- 
num. 

Conductivity  in  Matthiessen's  standard  scale. 

Cent. 

Fahr. 

1.00 

0.99 

0.98 

0.97 

0.96 

0.62 

0 
10 
20 

32 
50 
68 

19.2 
19.9 
20.7 

19.4 
20.1 
20.9 

19.6 
20.3 
21.  1 

19.8 
20.5 

21.3 

20.0 
20.8 
21.6 

30.9 
32.2 
33.4 

30 
40 
50 

86 
104 
122 

21.5 
22.3 
23.1 

21.7 
22.5 
23.3 

21.9 
22.7 
23.6 

22.2 
23.0 
23.8 

22.4 
23.2 
24.1 

34.7 
36.0 
37.3 

For  copper  or  aluminum  of  other  conductivity,  divide  value  under  1.00  by  given  conduc- 
tivity. 


CHAPTER  II. 
DISTRIBUTION  FOR  DIRECT-CURRENT  RAILWAYS. 

9.  Introduction.  —  Electric   railways  almost  always  use  the 
track  rails  for  the  return  of  current.     Besides  this,  other  differ- 
ences between  circuits  for  railways  and  those  for  power  and  light- 
ing purposes  are  the  higher  voltage  employed  on  the  trolley,  the 
greater  per  cent  loss  allowed  in  the  line  and  the  moving  and 
variable  nature  of  the  loads. 

Due  to  the  track  rails  the  calculations  of  railway  feeders  and 
working  conductors  are  somewhat  complex  and  uncertain. 
However,  the  recent  character  of  bonding  has  made  the  calcula- 
tions more  reliable  by  giving  a  higher  and  more  permanent  value 
to  the  conductivity  of  the  grounded  side.  In  electric  railways 
such  as  the  open  conduit  and  the  double  trolley  systems,  the 
track  rails  are  not  used  for  the  return  current,  and  the  calcula- 
tions for  their  circuits  are  therefore  simpler  and  more  definite. 
Where  the  feeders  and  working  conductors  form  a  complete 
copper  circuit,  the  formulas  on  page  14  may  be  used. 

10.  Resistance  of  Rails.  —  Table  9,  page  24,  gives  the  equiva- 
lent copper  section  and  the  resistance  of  third  rails  or  track  rails 
at  20°  cent.,  for  various  values  of  relative  resistance  of  steel  to 
copper.      The  corresponding  value  for  any  number  of  rails  is 
found  by  multiplying  the  equivalent  copper  section,  or  by  divid- 
ing the  resistance,  by  the  number.     The  resistance  at  any  other 
temperature  is  found  by  multiplying  the  value  in  the  table  by  the 
temperature  factor  of  resistance  for  steel,  T1}  Table  7,  page  17, 
while  the  equivalent  section  of  copper  is  equal  to  the  tabular 
value  divided  by  7\;  or,  the  values  may  be  found  for  one  rail  for 
any  other  condition  of  relative  resistance  and  temperature  by 
means  of  the  following  formulas: 

Equivalent  cir.  mils  of  100  per  cent  copper 

=  125,000  X  Pounds  per  yard 
Tl  X  Relative  resistance 

10 


DISTRIBUTION  FOR  DIRECT-CURRENT  RAILWAYS        17 


Ohms  resistance  per  1000  feet 


Tt  X  Relative  resistance 
12.1  X  Pounds  per  yard 


Ohms  resistance  per  mile 


T \  X  Relative  resistance 
2.28  X  Pounds  per  yard 


(21) 


(22) 


Tl  =  Temperature  factor  of  steel.     Table  7,  page  17. 
As  an  example,  the  total  resistance  per  mile  at  30°  cent,  of  two 
65-lb.  track  rails  having  a  relative  resistance  of  13.2  is 

1.05  X  13.2 


2.28  X  65  X  2 


=  0.0468  ohm. 


Table  7.— Values  of  T  for  Steel. 


Temperature,  deg.  cent  

0 

10 

20 

30 

40 

50 

Temperature,  deg.  fahr  

32 

50 

68 

86 

104 

122 

Values  of  T,  

0.91 

0.96 

1.00 

1.05 

1.09 

1.14 

ii.  Parallel  Resistance  of  Rails  and  Feeders. — The  parallel 
resistance  at  any  temperature  of  copper  and  aluminum  feeders 
of  any  conductivity  with  rails  of  any  relative  resistance  is  easily 
calculated  by  the  following  method: 

10,350 


Ohms  resistance  per  1000  feet  = 


Ohms  resistance  per  mile 


T,       T, 
54,700 


(23) 


(24) 


+ 


. 

r, 


Sl  —  Total  equivalent  cir.  mils  of  copper  in  rails.  Table  9,  page  24. 

*S2  =  Total  cir.  mils  in  copper  feeders. 

$3  =  Total  cir.  mils  in  aluminum  feeders. 

7\  =  Temperature  factor  of  steel.  Table  7,  page  17. 

T2  =  Temperature  factor  of  copper.     Table  1,  page  6. 

T3  =  Temperature  factor  of  aluminum.     Table  1,  page  6. 

Thus  the  resistance  per  mile  at  30°  cent,  of  two  80-lb.  rails 
of  relative  resistance  of  12  in  parallel  with  one  500,000-cir.  mil 


18  TRANSMISSION    CALCULATIONS 

copper  feeder  of  97  per  cent  conductivity  and  one  500,000-cir. 
mil  aluminum  feeder  of  62  per  cent  conductivity  is 

833^000  X  2       500,000      500,000  = 
1.05  1.069  1.674 

12.  Negative    Conductors.  —  The    track    rails    and    negative 
feeders  carry  the  return  current.     The  size  of  rails  is  fixed  by 
conditions  other  than  electrical  conductivity  and  usually  give  a 
total  resistance  much  below  that  of  the  positive  side.     Electro- 
lytic conditions  may  require  that  negative  feeders  be  connected 
to  the  rails  at  certain  points,  but  otherwise  the  rails  generally 
have  ample  conductivity  without  any  copper  reinforcement. 

The  section  of  negative  conductors  may  be  based  on  a  maxi- 
mum allowable  drop  in  the  return.  The  size  of  the  negative 
feeder  is  found  by  subtracting  the  equivalent  copper  section  of 
the  rails  from  the  total  circular  mils  required.  The  additional 
resistance  of  bonds  rnay  be  included  by  increasing  the  true 
relative  resistance  of  the  rail  to  an  apparent  value.  As  an 
example  of  the  determination  of  the  size  of  negative  feeder, 
suppose  that  S  in  formula  (25)  on  page  26  should  come  out 
2,000,000  cir.  mils  for  the  negative  side  of  a  circuit  for  which 
the  track  is  to  consist  of  two  70-lb.  rails  of  an  apparent 
relative  resistance  of  14  (including  bonding).  The  required 
feeders  in  parallel  with  the  track  should  have  2,000,000  — 
625,000  X  2,  or  750,000  cir.  mils  of  copper  of  100  per  cent  con- 
ductivity. Based  on  Table  8,  page  20,  this  is  equivalent  to 
773,000  cir.  mils  of  copper  of  97  per  cent  conductivity,  or  1,210,000 
cir.  mils  of  aluminum  of  62  per  cent  conductivity. 

13.  Positive  Conductors.  —  The  positive  conductors  consist  of 
auxiliary  feeders  in  parallel  with  trolleys  or  third  rails.     Trolley 
wires  vary  in  size  from  Nos.  0  to  0000,  while  third  rails  usually 
range  from  60  to  100  pounds  per  yard.     The  drop  in  the  positive 
conductors  is  found  by  subtracting  the  calculated  negative  drop 
from  the  total  that  is  allowed.     Then  for  a  third-rail  system, 
the  method  is  exactly  like  the  determination  of  negative  feeders 
in   paragraph  12;    while   for   systems   using   trolley   wires,  the 
total  section  of  positive  conductors  is  calculated  from  formula 
(25),  page  25.     The  size  of  the  auxiliary  feeder  is  given  by  the 
difference  between  the  total  circular  mils  required  and  that  of 
the  trolley  wire. 


DISTRIBUTION  FOR   DIRECT-CURRENT  RAILWAYS        19 

The  temperature  is  included  in  the  calculation  of  the  circular 
mils  by  means  of  A  in  Table  10,  page  26.  For  conductivities  other 
than  100  per  cent,  the  value  of  S  in  formula  (25)  is  divided  by 
the  given  conductivity.  Thus  if  S  comes  out  1,000,000  cir.  mils 
of  100  per  cent  copper,  the  required  section  of  copper  of  97  per 


1  ooo  000 

cent  conductivity  is     '       1      ,  or  1,030,000  cir.  mils,  as  may  be 
0.97 

noted  from  Table  8,  page  20. 

14.  Resistance  of  Circuit.  —  The  total  resistance  of  the  circuit 
is  obtained  by  adding  the  resistance  of  the  grounded  side  and 
the  overhead.     Where  there  are  no  negative  feeders,  the  resist- 
ance of  the  rails  may  be  taken  directly  from  Table  9,  page  24, 
and  if  no  positive  feeders  are  used,  the  resistance  of  the  trolleys 
is  given  in  Table  3,  page  7.      Thus  the  resistance  per  mile  at 
20°  cent,  of  a  single-track  road  having  two  60-lb.  rails  of  an 
apparent  relative  resistance   of  14  (including   bonding)   and  a 
No.  00  trolley  of  100  per  cent  conductivity  is 

2^2  +  o  411  =  o.462  ohm. 

15.  Given  Items.  —  The  determination  of  the  proper  loading  on 
which  to  base  the  feeder  systems  is  usually  more  difficult  than 
the  calculation  of  the  feeders.     In  general  the  requirement  will 
be  for  a  maximum  drop  with  a  very  severe  condition  of  loading  or 
for  a  much  smaller  drop  with  a  distributed  loading  of  an  average 
value.     In  either  case  the  loads  and  their  positions  are  first 
settled  and  then  the  formulas  are  applied  successively  to  the 
separate  loads;  or,  where  the  feeder  system  is  uniform  through- 
out, one  determination  is  made,  using  the  total  of  the  loads  at 
their  center  of  gravity. 

In  case  the  feeder  system  is  known  the  total  loss  is  easily 
calculated  by  considering  the  loads  separately,  or  by  considering 
their  combined  effect  where  the  feeder  system  is  uniform 
throughout  the  area  of  loading. 

For  convenience  in  calculation  the  formulas  have  been  ex- 
pressed in  terms  of,  current.  The  per  cent  volt  loss  has  also 
been  given  in  terms  of  power,  so  that  no  preliminary  approxi- 
mation is  necessary  when  the  voltage  is  known  only  at  the  source. 

16.  Examples.  —  Although^  in  practice,  these  problems  occur 
in  a  great  many  different  forms,  the  application  of  the  formulas 

.will  be  clear  from  the  following  typical  examples. 


20  TRANSMISSION    CALCULA  TIONS 

Table  8.  — Equivalents  of  Copper  of  100  Per  Cent  Conductivity. 


Conductivity  in  Matthiessen's  Standard  Scale. 

1.00 

0.99 

0  98 

0.97 

0.96 

0.62 

1,000,000 
900,000 
800,000 
700,000 

1,010,000 
909,000 
808,000 
707,000 

1,020,000 
918,000 
816,000 
714,000 

1,030,000 
928,000 
825,000 
722,000 

1,040,000 
938,000 
834,000 
730,000 

1,640,000 
1,450,000 
1,290,000 
1,130,000 

600,000 
500,000 
400,000 
300,000 

606,000 
505,000 
404,000 
303,000 

612,000 
510,000 
408,000 
306,000 

618,000 
515,000 
412,000 
309,000 

625  :  000 
521,000 
417,000 
312,000 

984,000 
820,000 
656,000 
492,000 

For  other  sections,  divide  given  circular  mils  of  100  per  cent  copper  by  given  conductivity. 
For  other  conductivity,  divide  circular  mils  under  1 .00  by  given  conductivity. 
To  find  equivalent  circular  mils  of  100  per  cent  copper  in  a  wire,  multiply  given  circular 
mils  by  given  conductivity. 

Example  6.  —  Find  the  negative  drop  per  100  amperes  at 
30°  cent,  in  a  mile  of  track  consisting  of  two  90-lb.  rails  having 
a  relative  resistance  of  12.5  (including  bonding)  in  parallel 
with  97  per  cent  copper  feeders  of  0,  500,000,  750,000,  and 
1,000,000  cir.  mils,  respectively. 

From  Table  9,  page  24,  equivalent  cir.  mils  of  100  per  cent 
copper  in  two  rails  at  20°  cent.  =  1,800,000. 

From  Table  7,  page  17,  temperature  factor  for  steel  at  30° 
cent.,  Tl=  1.05. 

From  Table  1,  page  6,  temperature  factor  for  97  per  cent 
copper  at  30°  cent.,  T=  1.069.  Then  from  (24)  and  (28),  the 
following  values  are  derived: 


Rails. 

.  Copper  Feeder. 

Resistance  per 
Mile. 

Drop  per  Mile  per 
100  Amperes. 

Lb.  per  Yd. 

Cir.  Mils. 

Ohm. 

Volts. 

2-90 

0 

0.0319 

3.19 

2-90 

1-500,000 

0.0251 

2.51 

2-90 

1-750,000 

0.0226 

2.26 

2-90 

1-1,000,000 

0.0206 

2.06 

Owing  to  the  comparatively  large  section  of  the   rails,   the 
combined  resistance  is  reduced  but  18  per  cent  for  an  increase 


DISTRIBUTION  FOR  DIRECT-CURRENT  RAILWAYS         21 

in  the  negative  feeder  from  500,000  to  1,000,000  cir.  mils,  equal 
to  an  increase  of  100  per  cent. 

Example  7.  —  For  a  temperature  of  40°  cent.,  determine  the 
line  voltage  at  successive  cars  which  take  100  amperes  each  at 
respective  locations  of  500,  2000,  3000,  and  5000  feet  from  a 
power  station  generating  550  volts,  if  the  circuit  consists  of  four 
80-lb.  track  rails  having  a  relative  resistance  of  14  (including 
bonding),  and  two  No.  00  trolleys  of  97  per  cent  conductivity 
in  parallel  with  one  500,000  cir.  mil  aluminum  feeder  of  62 
per  cent  conductivity. 

From  Tables  7  and  9,  the  resistance  of  the  tracks  is, 

Resistance  per  1000  feet  =  °-0145  X  i-09  =  0.00395  ohm. 

From  (23)  and  Tables  1  and  3,  the  resistance  of  the  overhead  is, 
Resistance  per  1000  feet 


10350 


133,100  X  2      500,000 
1.11  1.738 


=  Q  01962  ohm< 


Hence  the  total  resistance  per  1000  feet  of  road  is 
0.00395  +  0.01962  =  0.02357  ohm. 

1st  Car  —  Total  drop          =  0.02357  X  400  X0.5  =  4.7  volts. 

—  Line  volts  =  550  -  4.7  =  545.3. 

2d    Car  —  Additional  drop  =  0.02357  X  300  X  1  .500  =  10.6  volts. 

—  Line  volts  -  545.3  -  10.6  -  534.7. 
3d    Car  —  Additional  drop  =  0.02357  X  200  X  1  .000  =  4.7  volts. 

—  Line  volts  =  534.7  -  4.7  =  530. 

4th  Car  —  Additional  drop  =  0.02357  X  100X2.000  =  4.7  volts. 

—  Line  volts  =  530  -  4.7  =  525.3. 

In  the  above  typical  problem,  the  resistance  of  the  grounded 
side  is  but  20  per  cent  of  the  resistance  of  the  overhead. 

Example  8.  —  A  single-track  road  with  two  75-lb.  rails  having 
a  relative  resistance  of  13  plus  10  per  cent  for  bonding  (apparent 
relative  resistance  =  14.3),  is  to  supply  four  cars  with  150 
amperes  each  when  located  at  0.8,  1.0,  1.5  and  2.5  miles  from  a 
substation  generating  550  volts.  If  the  minimum  line  e.m.f. 
is  to  be  400  volts  at  50°  cent.,  what  size  copper  feeder  of  97  per 
cent  conductivity  should  be  in  parallel  with  a  No.  00  trolley 
of  96  per  cent  conductivity? 


22  TRANSMISSION    CALCULATIONS 

From  (22)  and  Table  7,  page  17, 

Resistance  per  mile  of  two  rails  =    1<14  X  14'3    =  0.0477  ohm. 

2.28  X  75  X  2 

Total  drop  in  track  =  0.0477  X  150  (0.8  +  1.0  +  1.5  +  2.5)  - 
41.5  volts. 

Allowable   drop  in   overhead  =  150  -  41.5=  108.5  volts. 

From  (25),  in  which  IL  is  the  sum  of  products  of  each  load 
multiplied  by  its  distance  from  source,  and  from  Table  10,  page  26, 

s  =  61,100  X  150(0.8  +  1.0  +  1.5  +W  =  490>000  ci,  mils 

lOo.O 

of  100  per  cent  copper. 

Based  on  Table  8,  page  20,  the  equivalent  section  of  100  per  cent 
copper  in  trolley  is 

133,100  X  0.96=  128,000  cir.  mils. 
Required  section  of  100  per  cent  copper  in  feeder 

-  490,000  -  128,000  =  362,000  cir.  mils. 
Required  section  of  97  per  cent  copper  in  feeder 

=  5^2,220  ,  373,000  cir.  mils. 
0.97 

Example  9. —  Find  the  size  of  a  third  rail  of  relative  resistance 
7.5  (including  bonding)  required  to  start  two  cars  taking  1000 
amperes  at  a  point  midway  between  substations  8  miles  apart, 
if  the  drop  at  20°  cent,  is  to  be  25  per  cent  of  the  600  volts  at 
rotaries.  Each  track  rail  is  to  weigh  80  Ib.  per  yard  and  to 
have  a  relative  resistance  of  13  (including  bonding). 

Distance  from  either  substation  to  cars  =  f  =  4  miles. 

Current  from  each  substation  =  ^-^  =  500  amperes. 

Total  allowable  drop  =  600  X  0.25  =  150  volts. 

From  Table  9,  page  24,  resistance  per  mile  of  two-track  rails 
=  0.0713  =  003565  ohm 

Drop  in  track  =  0.03565  X  500  X  4  =  71.3  volts. 
Allowable  drop  in  third  rail  =  150  -  71.3=  78.7  volts. 
From  (25),  in  which  A  from  Table  10,  page  26  =  54,700, 
s  =  54,700  x  500  XJ= 


DISTRIBUTION  FOR  DIRECT-CURRENT  RAILWAYS        23 

From  Table  9,  page  24,  this  section  corresponds  to  an  85-lb. 
rail.  Then  from  above  and  Table  9, 

Total  resistance  per  mile=  R  =  0.03565  +  0.0387  =  0.07435  ohm. 

From  (28),  total  drop  =  0.07435  X  500  X  4  =  148.7  volts. 

From  (37),  voltage  at  cars  =  600  -  148.7=  451.3  volts. 

From  (40),  drop  in  per  cent  of  volts  at  load  is 

V  =  148.7^-0.01  X  451.3  =  33.0  per  cent. 

From  (41),  drop  in  per  cent  of  volts  at  substation  is 
70=  148.7-r-0.01  X  600  =  24.8  per  cent. 

Watt  loss.  —  From  above,  R  =  0.07435  ohm. 

From  (35),  p  =  0.07435  X  (500)2  X  8 

=  148,700  watts. 

Example  10.  —  Calculate  the  size  of  positive  feeders  at  30°  cent, 
required  for  a  uniformly  distributed  load  of  600  amperes  per  mile 
between  two  substations  6  miles  apart  with  an  average  loss  of 
5  per  cent  of  the  650  volts  generated,  if  there  are  to  be  eight 
100-lb.  track  rails  and  four  70-lb.  third  rails  having  relative 
resistances  (including  bonding)  of  12.5  and  8.0,  respectively. 

Since  the  average  loss  is  to  be  5  per  cent,  the  maximum  is  10 
per  cent  or  65  volts.  The  result  is  equivalent  to  the  total  load 
concentrated  at  a  point  one-quarter  the  distance  between  sub- 
stations from  either  one. 

fiOO  V  fi 

Total  load  per  substation  =         ^     =  1800  amperes. 

£ 

Center  of  gravity  of  load  =  f  =  2  miles  from  either  sub- 
station. 

From  Tables  7  and  9,  resistance  per  mile  of  four  tracks 

=  0.0548  X  1.05-5-8  =0.00719  ohm. 
Drop  in  track  =  0.00719  X  1800  X  2  =  25.9  volts. 
Allowable  positive  drop  =  65  -  25.9=  39.1  volts. 
From  (25),  5  =  57,400  X  1800  X  2-4-39.1  =  5,280,000  cir.  mils 
of  100  per  cent  copper. 

From  Tables  7  and  9,  third-rail  section  =1,090,000  X  4-*- 1.05 

=  4,150,000  cir.  mils  of  100  per  cent  copper. 
Required  section  of  100  per  cent  copper  feeder 

=  5,280,000  -  4,150,000  =  1,130,000  cir.  mils. 
Required  section  of  97  per  cent  copper  feeder 

=  1,130,000-4-0.97  =  1,165,000  cir.  mils. 
Required  section  of  62  per  cent  aluminum  feeder 
=  l,130,000-s-0.62  =  1,823,000  cir.  mils. 


TRANSMISSION    CALCULATIONS 


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DISTRIBUTION  FOR  DIRECT-CURRENT  RAILWAYS        25 


Formulas  for  D.  C.  Railway  Circuits. 


Required  Items. 

Positive  or  Negative  Alone. 

Positive  and  Negative  Total. 

Circular  mils 
or 
Total  resistance. 

A  from  Table  10,  page  26. 

.-•*...« 

S  =  —  .    .  (25) 
v 

Volt  drop. 

.-^..(») 

See  Par.  14  ,  page  19. 

V  =  EIL.    .    . 

(28) 

Per  cent  volt  drop 
or 
Per  cent  power  loss 
from  amperes. 

y        AIL        ,.>9) 

y                RIL 

(30) 

OOlSe 

0.01  Se 

y         AIL       ,31) 

y                     RIL 

(32) 

0      0.01  Se0 

0      0.01  Se0  ' 

Per  cent  volt  drop 
from 
watts  at  load. 

y  „,        AwL       ,33, 

y,n           RwL 

(34) 

0        O.OlSe* 

0.01  Sej' 

TO  from  Table  31,  page  72. 

Watt  loss. 

T  from  Table  1  or  7.    R  from  Table  3,  page  7. 

p  =  TRIZL  =  0.01  Vw  =     °-01T^      .    . 

*                 0 

(35) 

Volts  at  source. 

e0=  e  +  v  =  e(l  +0 

01  V}                e 

(36) 

l)      1-0.01F0' 

Volts  at  load. 

.-^-.-n^-^-o-oi^cw) 

Watts  at  source. 

7/1              7/J   4     T)           M(\      \      0   OIF1)    —                    W 

(38) 

W»                                                  1-0.01F/ 

Amperes. 

z      w                  w                  wn 

(39) 

e       e0(l-0.01F0)       e0 

When  given  F,  find  v  from  v  =  0.01  Fe  =  0.01  Ve0  /  (1  +  0.01  F)   .    .    . 
When  given  F0,  find  v  from  v  =  0.01F0e0  =  0.01F0e  /  (1  -  0.01F0)    .    . 
When  given  w,  find  I  from  (39)  above. 

(40) 
(41) 

NOTATION. 

^  =  Resistance  factor.     Table  10,  page  26.        S  =  Equivalent  cir.  mils  of  copper  of  100  per 
e  =  Volts  at  load.  cent  conductivity. 

e0  =  Volts  at  source.  T  =  Temperature  factor.    Table  1,  page  6,  or 

/  =  Total  current  at  source,  in  amperes.  Table  7,  page  17. 

L  —  Distance  from  source  to  load,  in  miles  or     V  =  Volt  loss  in  per  cent  of  volts  at  load. 

1000  feet.  ^0  =  Volt  loss  in  per  cent  of  volts  at  source. 

p  =  Power  loss  in  watts.  Vn'"  =  Volt  loss  factor.    Table  31,  page  72. 

/  Ohms  resistance  per  mile  for  L  in  miles.      v  =  Volt  drop. 
R  =  ]  Ohms  resistance  per  1000  feet  for  L  in      w  =  Watts  at  load. 

'     1000  feet.  w'o  =  Watts  at  source. 


26 


TRANSMISSION    CALCULATIONS 


Table  10. —  Values  of  A  for  Wires  and  Rails. 


Temperature  in 
degrees. 

Wires. 

Rails. 
With    or     without 
feeders. 

Cent 

Fahr. 

For  L 
in.  Miles. 

ForL 
in.  1000  Ft 

For  L 

in.  Miles. 

For  I, 
in.  1000  Ft. 

0 

32 

50,600 

9,580 

49,800 

9.420 

10 

50 

52,700 

9,970 

52,500 

9,950 

20 

68 

54,700 

10,350 

54,700 

10,350 

30 

86 

56,800 

10,740 

57,400 

10,870 

40 

104 

58,900 

11,140 

59,600 

11,280 

50 

122 

61,100 

11,560 

62,400 

11,800 

The  above  values  for  wires  are  for  copper  of  100  per  cent  conductivity  in  Matthiessen's 
standard  scale. 

The  above  values  for  rails  are  for  equivalent  circular  mils  of  copper  of  100  per  cent 
conductivity  in  Matthiessen's  standard  scale. 


PART  II. 

ALTERNATING-CURRENT    TRANSMISSION 

BY    MEANS    OF 
OVERHEAD,  UNDERGROUND   AND  INTERIOR   WIRES 

FOR    PURPOSES    OF 
LIGHT,    POWER    AND    TRACTION 


27 


ALTERNATING-CURRENT   TRANSMISSION 


CHAPTER   III. 
ALTERNATING-CURRENT   TRANSMISSION    BY   OVERHEAD   WIRES 

17.  Introduction.  —  The  methods  in  common  use  for  calcu- 
lating alternating-current  transmission  lines  from  the  volt  loss 
are  either  indirect  or  inaccurate.  This  condition  results  from 
the  fact  that  it  is  impossible  to  devise  an  exact  formula  for  any 
alternating-current  system  which  will  directly  indicate  the  size 
of  wire  required  for  the  transmission  of  a  given  amount  of  power 
with  a  given  volt  loss.  Methods  of  the  indirect  class  require 
that  the  size  wire  be  known,  and  hence  are  trial  methods. 
Approximations  of  the  second  class  may  be  sufficiently  close 
under  certain  conditions  but  give  wholly  erroneous  results  for 
other  practical  cases. 

The  essential  feature^pLthe  desirable  method  is  that  it  should 
directly  determine  for  any  system  and  frequency,  the  size  of 
wire,  power  loss  and  power-factor  at  generator  when  given  the 
volt  loss,  length  of  line,  power  received,  load  power-factor, 
voltage  at  generator  or  load  and  distance  between  wires.  The 
following  method  accomplishes  this  result  with  commercial 
accuracy  over  a  sufficiently  wide  range  of  conditions  to  cover 
almost  air  practical  cases. 

One  familiar  method  of  calculating  alternating-current  lines 
assumes  that  the  impedance  volts  equals  the  line  loss,  or  in  Figs. 
1  and  2  AB  =  AC.  Several  other  methods  in  common  use 
assume  that  the  line  loss  equals  the  projection  of  the  impedance 
volts  on  the  vector  of  delivered  voltage.  The  errors  of  these 
approximations  in  terms  of  the  actual  volt  loss  are  shown  in 
Table  11,  for  copper  wires  on  36-inch  centers  and  a  true  volt 
loss  of  10  per  cent  of  the  generated  volts. 

The  errors  for  leading  power-factors,  larger  wires  or  greater 
spacings  exceed  those  shown  in  Table  11,  and  it  is  evident 
then,  that  both  methods  may  lead  to  erroneous  results.  The 

29 


30 


TRANSMISSION    CALCULATIONS 


first  assumption  gives  wires  which  are  too  large,  while  the 
other  assumption  gives  wires  too  small.  By  stating  the  errors 
of  Table  11  in  terms  of  the  calculated  values,  the  percentages 
shown  would  be  greatly  decreased  for  the  first  assumption  and 
similarly  increased  in  the  second  case. 

18.  Outline  of  Method.  —  All  items  which  depend  on  the 
frequency,  size  of  wire,  load  power-factor  and  wire  spacing  were 
grouped  into  one  term,  which  is  denoted  by  M  and  called  the 
wire  factor.  All  other  variables  that  determine  the  size  of  wire 


Fig.  2 


were  grouped  into  the  second  member  of  the  equation  of  M. 
Then  after  introducing  the  proper  transmission  factors,  the 
equation  took  the  form  shown  in  the  table  of  formulas  opposite 
the  size  of  wire.  The  remaining  formulas  are  simple  derivations 
either  from  the  equation  of  M  or  from  well  known  relations. 

The  vector  diagram  from  which  the  fundamental  equation  was 
derived  is  shown  in  Fig.  1  for  lagging  current  and  in  Fig.  2  for 
leading  current,  wherein, 


ALTERNATING-CURRENT  TRANSMISSION 


31 


01  =  Current  at  load. 
OA  =  Volts  at  load. 
AD=  Resistance  volts. 
DB  =  Reactance  volts. 
AB=  Impedance  volts. 


OB  —  Volts  at  source. 

AC=  OB  -  OA  =  Volt  loss. 

<p  =  Power-factor  angle  of  load. 

0    =  Power-factor  angle  of  line. 

(f>Q  =  Power-factor  angle  at  source. 


19.  Range  of  Application.  —  The  formulas  apply  for  all  volt 
losses  less  than  20  per  cent  of  the  generated  voltage  or  25  per 
cent  of  the  voltage  at  load,  and  for  a  sufficient  range  of  wire 
sizes  to  cover  the  usual  and  unusual  cases  in  practice.  At  high 
standard  frequencies  and  for  power-factors  near  100  per  cent 
the  values  of  M  for  the  largest  wires  were  omitted  in  order  to 
confine  the  maximum  possible  error  within  the  limits  prescribed 
in  Table  12,  page  32. 

Table  n.  —Error  in  Per  Cent  of  True  Volt  Loss. 


Size  of 
Wire. 
A.W.G. 

25  Cycles  per  Sec. 

60  Cycles  per  Sec. 

125  Cycles  per  Sec. 

Lagging  Power  Factor  of  Load  in  Per  Cent. 

100 

90 

80 

100 

90 

80 

100 

90 

80 

Assuming  Impedance  Volts  =  Volt   Loss. 

0000 
0 
4 

36 
11 
1 

4 
0 
1 

0 
0 

7 

110 
53 

12 

29 
10 
0 

13 
2 
0 

200 
125 

44 

55 
33 

7 

29 
16 

1 

Assuming  Projection  of  Impedance  Volts  =  Volt  Loss. 

0000 
0 
4 

5 
1 
0 

0 
0 
0 

0 
0 
0 

19 
8 
1 

4 
1 
0 

1 
0 
0 

43 
22 
6 

8 
4 
0 

4 
1 
0 

20.  Maximum  Error.  —  As  previously  stated,  any  direct 
method  must  be  approximate.  However,  all  errors  introduced 
by  the  following  method  are  easily  within  conservative  limits 
and  usually  negligible.  In  determining  the  size  of  wire  from  the 
volt  loss  the  maximum  error  is  about  10  per  cent  of  the  wire 
factor  M,  which  is  usually  less  than  the  difference  of  M  for 
successive  wires  in  commercial  sizes.  The  maximum  error 
occurs  near  0  per  cent  and  20  per  cent  volt  losses  but  gradually 


32 


TRANSMISSION    CALCULATIONS 


reduces  to  a  zero  error  near  10  per  cent  loss.  The  point  where  the 
formula  for  M  should  be  correct  was  arbitrarily  fixed  at  10  per 
cent  volt  loss,  since  this  is  about  the  mean  value  in  practice, 

After  the  wire  is  determined,  the  actual  per  cent  drop  is  calcu- 
lated. In  order  to  minimize  its  error  and  that  of  the  remaining 
items,  each  column  in  the  tables  of  M  has  been  divided  into  a 
maximum  of  three  sections.  It  will  be  observed  that  any 
value  of  M  is  found  between  the  section  letters  a  and  b,  b  and  c, 
or  c  and  d,  depending  upon  the  size  of  wire,  frequency,  spac- 
ing and  power-factor. 

The  maximum  errors  in  the  calculations  are  shown  in  Table 
12  below.  These  errors  occur  near  0  per  cent  and  20  per  cent 
volt  losses  but  gradually  reduce  to  zero  near  10  per  cent  loss. 
In  practical  problems  the  calculated  value  of  M  is  most  often 
between  the  section  letters  c  and  d,  and  therefore  the  errors  of 
the  remaining  calculations  are  negligible. 

It  should  be  observed  that  the  values  in  Table  12  are  expressed 
in  per  cent  of  the  true  result.  Thus  for  an  error  of  2  per  cent 
with  a  true  volt  loss  of  5  per  cent,  the  calculated  value  would 
be  no  greater  than  5.1  and  no  less  than  4.9  per  cent. 

Table  12.  —  Maximum  Errors  in  Per  Cent  of  True  Values 
at  20°  Cent. 


Calculated  Items. 

For  M  in  Section 

a--b. 

b-c. 

c  —  d. 

Volt  loss  or  per  cent  volt  loss 

6.0 
1.5 
0.0 

3.0 
0.8 
0.0 

1.5 
1.3 

2.3 

4.0 
1.0 
0.0 

2.0 
0.5 
0.0 

!.o 

1.0 
1.5 

2.0 
0.5 
0.0 

1.0 
0.3 
0.0 

0.5 
0.5 
0.8 

Xilovolts  at  load  or  source 

Power  loss  or  kilowatts  at  source,  when  given  E     .  . 

Power  loss  or  per  cent  power  loss,  when  given  Z?0  
Kilowatts  at  source  when  given  EG.  ...                       .       . 

Amperes  per  wire,  when  given  E       

Amperes  per  wire,  when  given  En     .                     .            ... 

Power-factor  at  source,  when  given  E  
Power-factor  at  source,  when  given  E0  

21.  Transmission  Systems.  —  Systems  for  alternating-current 
transmission  are  either  one,  two  or  three-phase.  These  results 
are  applicable  to  one  and  three-phase  systems,  and  to  the  two- 
phase  system  with  either  three  or  four  wires  of  equal  size. 


ALTERNATING-CURRENT   TRANSMISSION  33 

Accurate  formulas  cannot  be  given  for  ,the  two-phase  three-wire 
system  where  the  size  of  the  common  wire  is  different  from  the 
others.  However,  the  result  may  be  .approximated  by  calcu- 
lating for  three  equal  wires  and  then  making  the  proper  allow- 
ance for  the  larger  cross  section  of  the  common  lead.  (See 
Example  16,  page  44.)  In  the  three-phase  four- wire  system 
the  neutral  wire  carries  no  current  when  the  system  is  balanced, 
and  hence  the  results  are  exactly  similar  to  those  for  the  three- 
phase  system  with  three  wires. 

22.  Balanced  Loads.  —  A  balanced  load  is  assumed  in  all  the 
formulas.     In   practice   such   is   often   not   the   case,    although 
the  variation  from  a  balanced  condition  is  usually  small.     The 
effect  of  unbalancing  is  to  alter  the  voltage  in  proportion  to  the 
amount  of  unbalancing.     The  other  items  are  also  affected,  but 
seldom  is  the  discrepancy  of  any  practical  importance. 

23.  Temperature.  —  All   results   have   been   calculated  for  a 
temperature  of  20°  cent,  or  68°  fahr.     The  value  of  M  varies 
directly  with  the  temperature,  but  it  also  depends  to  a  lesser 
degree  on  other  conditions.     However,  as  the  error  of  commission 
is  much  less  than  the  error  of  omission,  the  values  of  A  in  Table 
16,  page  52,  have  been  calculated  for  various  temperatures. 

24.  Specific  Conductivity.  —  The  calculations  have  been  made 
for  a  specific  conductivity  of  100  per  cent  for  copper  and  62  per 
cent  for  aluminum.     The  value  of  M  varies  inversely  with  the 
specific   conductivity  but  also   depends  somewhat  upon  other 
conditions.     However,  sufficiently  accurate  results  are  obtained 
by  including  the  proper  conductivity  of  the  conductor  by  the 
method  shown  in  Table  16. 

25.  Solid  and   Stranded   Conductors.  -     Wires   smaller   than 
No.  0  have  been  considered  solid,  while  the  larger  sizes  have  been 
taken  stranded.     However,  since  the  inductance  of  a  stranded 
wire  is  between  that  of  a  solid  wire  of  the  same  cross  section  of 
metal  and  one  of  the  same  diameter,  the  discrepancy  in  the  final 
results  for  any  ordinary  variation  from  the  assumed  conditions  is 
not  appreciable. 

26.  Skin   Effect.  —  Owing   to    a   decreasing   current    density 
toward  the  center  of  wires  carrying  alternating  currents,   the 
effective   resistance  is   increased    in   direct    proportion    to    the 
product  of  its  cross  section  and  the  frequency.     In  the  usual 
sizes  of  transmission  wire  the  effect  is  negligible,  and  even  in  the 


34  TRANSMISSION  CALCULATIONS 

largest  sizes  shown  for.  the  higher  frequencies  the  maximum 
additional  resistance  is  less  than  5  per  cent,  which  value  decreases 
very  rapidly  with  the  size  of  wire. 

27.  Wire  Spacing.  -  -  Inductance  increases  directly  with  the 
distance  from  center  to  center  of  the  wires.     However,  the  effect 
on  the  impedance  of  the  line  for  large  variations  in  the  spacing 
is  not  great,  even  at  the  higher  commercial  frequencies.      In 
order  to  cover  all  cases  in  practice,  the  wire  factor  has  been 
calculated  for  three  different  spacings  at  each  frequency.     In 
general  the  results  for  wires  on  18  and  60-inch  centers  will  be 
sufficiently  accurate  for  spacings  less  than  27  inches  and  greater 
than  48  inches,  respectively,  while  the  values  for  36  inches  will 
cover  spacings  from  27  to  48  inches.     But,  where  greater  accuracy 
is  desired  for  spacings  other  than  shown,  the  values  of  M  may  be 
readily  interpolated  from  the  tables. 

28.  Arrangement  of  Wires.  —  It  is  assumed  in  the  two  and 
three-phase  systems  with  three  wires  that  the  conductors  are 
placed  at  the  three  corners  of  an  equilateral  triangle.     For  other 
arrangements,  with  the  wires  properly  transposed,  little  error  is 
introduced  by  taking  the  distance  from  center  to  center  of  the 
wires  as  the  average  of  the  distances  between  wires  1  and  2, 
2  and  3,  and  1  and  3.     In  the  two-phase  four-wire  system  the 
distance  between  wires  of  the  same  phase  should  be  used. 

29.  Frequency.  —  The   tables    have   been   calculated   for   all 
standard  frequencies.     The  values  for  15  cycles  per  second  are 
necessary  for  the  design  of  single-phase  railway  systems,  while 
25,  40,  60,  and  125  cycles  per  second  cover  the  remaining  systems 
of  transmission  for  purposes  of  light,  power,  and  traction.     How- 
ever, where  an  odd  frequency  is  to  be  employed,  the  required 
value  of  M  may  be  interpolated  from  the  tables. 

30.  Multiple  Circuits.  —  Where  circuits  are  in  parallel  from 
the  source  to  receiver,  the  load  should  be  proportioned  between 
them  and  each  line  calculated  separately. 

31.  Current-carrying     Capacity.  —  The  current-carrying  capa- 
city in  amperes  per  wire  is  shown  in  Table  4,  page  8,  for  both 
copper  and  aluminum.     The  values  in  the  table  are  based  on  a 
temperature   rise   of   40°  cent,  or   72°  fahr.,  but   the  current- 
carrying  capacity  for  any  temperature  elevation  may  be  found 
from  formula  (4),  page  9. 

In  general,  overhead  transmission  lines  are  of  sufficient  length 


ALTERNATING-CURRENT   TRANSMISSION  35 

to  insure  the  necessary  current-carrying  capacity  when  the 
conditions  of  volt  loss  are  met.  However,  it  is  desirable  to  note 
from  Table  4  that  such  is  the  case,  after  each  determination  of 
wire. 

32.  Transmission  Voltage.  —  The  voltage  is  taken  between 
wires,  either  at  the  load  or  source.     The  term  "source"  designates 
the  generator,  the  secondary  terminals  of  step-up  or  step-down 
transformers,  or  a  certain  point  of  the  circuit  from  which  the 
calculation  is  made.     In  general  the  voltage  at  the  source  is 
given,  although  in  some  cases  of  transmission  to  a  single  center 
of  distribution,  the  voltage  at  the  load  is  specified.     The  formu- 
las have  been  stated  in  terms  of  both  voltages  in  order  to  cover  all 
cases  without  any  preliminary  approximation.     For  convenience 
the  voltage  has  been  expressed  in  kilovolts,  thousands  of  volts. 
In  a  two-phase  four-wire  system  the  voltage  between  the  wires 
of  the  same  phase  is  specified. 

With  lagging  current  the  voltage  at  the  load  is  always  less 
than  the  voltage  at  the  generator.  With  leading  current  the 
voltage  at  load  is  less  when  the  sum  of  the  power-factor  angles 
of  load  and  line  is  less  than  90  degrees,  but  it  gradually  becomes 
greater  than  the  voltage  at  the  generator  as  their  sum  increases 
above  90  degrees. 

33.  Volt  Loss.  —  In  an  alternating-current  system  the  volt 
loss  in  the  line  is  the  difference  between  the  voltages  at  the 
generator  and  at  the  load.     The  line  loss  is  always  less  than  the 
impedance  volts  and  almost  always  greater  than  the  projection 
of  the  impedance  volts  on  the  vector  of  delivered  volts.     It  may 
be  greater  or  less  than  the  resistance  volts.     (See  Figs.  1  and  2, 
page  30.) 

The  per  cent  volt  loss  may  be  expressed  in  terms  of  the  volts 
at  load  or  source.  However,  either  may  be  obtained  from  the 
other  by  means  of  the  simple  equations  shown  below  the  table 
of  formulas  on  page  50.  For  convenience  in  calculations  the 
per  cent  loss  is  expressed  as  a  whole  number. 

34.  Power  Transmitted.  —  The  power  transmitted  is  always 
specified  at  the  load,  and  is  usually  expressed  as  true  power  in 
kilowatts   or  as   apparent   power  in   kilo  volt-amperes.     Where 
the  load  is  given  in  amperes  the  equivalent  value  of  kilowatts 
may  be  obtained  from  equations  (59)  or  (60)  solved  for  W.    The 
values  of  B  in  Table   17,  page  52,  serve  to  make  the  formula 


36  TRANSMISSION  CALCULATIONS 

for    amperes    per   wire    applicable    to    all    systems    of    trans- 
mission. 

For  balanced  loads  in  one-phase,  two-phase  four-wire  or  three- 
phase  circuits  the  current  is  the  same  in  each  wire,  while  in  the 
two-phase  three-wire  system,  the  current  in  the  common  lead  is 
1.41  times  that  in  each  of  the  other  wires.  In  the  two-phase 
three-wire  system  /  represents  the  amperes  in  each  of  the  outer 
wires,  this  being  the  same  as  the  current  per  phase. 

35.  Power  Loss.  —  The  power  loss  in  any  system  depends  only 
upon  the  current  and  resistance.     The  per  cent  power  loss  may 
be  greater  or  less  than  the  per  cent  volt  loss.     It  is  less  than  the 
per  cent  volt  loss,  V,  when  the  power-factor  at  the  source  is  less 
than  the  power-factor  of  the  load,  and  is  greater  when  the  reverse 
is  true.     An  exception  to  this  statement  may  occur  when  capacity 
effects  in  the  line  are  included.    (See  Example  20,  page  48,  and 
Example  25,  page  69.)     The  power  at  the  source  is  the  sum  of 
the  power  at  load  and  the  power  loss  in  line. 

36.  Power-Factor.  —  The  power-factor  of  the  load  is  always 
given.     The  values  of  M  have  been  calculated  from  95  per  cent 
lead  to  75  per  cent  lag,  for  all  frequencies  except  125  cycles  per 
second,  giving  a  range  sufficient  to  cover  almost  all  practical 
cases  of  transmission.     The  power-factor  at  the  load  should  be 
known  accurately,  as  the  value  of  M  varies  greatly  with  it. 

The  power-factor  at  the  source  depends  upon  the  power-factor 
at  the  load,  the  per  cent  power  loss  and  the  per  cent  volt  loss. 
For  current  leading  at  the  load,  the  power-factor  is  less  leading 
at  the  source  than  at  the  load,  and  even  may  become  lagging  at 
the  source.  (See  Example  24,  page  68.)  Lagging  current  at  the 
load  is  always  accompanied  by  lagging  current  at  the  source. 
The  power-factor  at  'the  source  is  lower  than  the  power- 
factor  of  the  load  when  the  volt  loss,  V,  is  a  larger  percentage 
than  the  power  loss,  P,  and  it  is  higher  at  the  source  when  the 
reverse  is  true.  This  statement  may  be  modified  by  capacity 
effects  in  the  line.  (See  Example  20,  page  48,  and  Example  25, 
page  69.) 

37.  Wire  Factor.  —  The  wire  factor,  M,  depends   upon  the 
impedance  of  the  wires,  and  on  the  power-factor  angles  of  the 
load  and  line.     The  values  have  been  calculated  for  copper  and 
aluminum  at  all  standard  frequencies,  for  a  range  of  sizes  and 
load  power-factors  sufficient  to  cover  all  practical  cases  likely 


ALTERNATING-CURRENT  TRANSMISSION  37 

to  arise  in  transmission  design.  Tables  18  to  22  give  the  results 
for  copper,  and  Tables  23  to  27  for  aluminum.  Those  values 
which,  under  certain  conditions,  might  lead  to  errors  greater 
than  previously  specified  have  been  omitted. 

38.  Given  Items.  —  The  given  problem  may  be  of  two  kinds; 
either  the  line  loss  is  specified  and  the  size  of  wire  is  required,  or 
the  size  of  wire  is  specified  and  the  line  loss  is  required.     In  both 
cases  the  additional  items  to  be  given  are  system  of  transmission, 
conductor  metal,  temperature,  frequency,  wire  spacing,  distance 
of  transmission,  power  delivered,  power-factor  of  load,  and  the 
voltage  between  wires  at  the  source  of  load. 

39.  Size  of  Wire.  —  The  size  of  wire  is  generally  determined 
from  the  per  cent  volt  loss,  although  in  some  instances  it  is  fixed 
by  the  condition  of  per  cent  power  loss.     Formulas  are  given  for 
both  methods  of  procedure. 

When  the  per  cent  volt  loss  is  given,  the  value  of  A  is  taken 
from  Table  16,  page  52,  for  the  given  system  of  transmission, 
temperature  and  specific  conductivity.  From  Table  15,  page  51, 
V  or  VQ  is  found,  depending  upon  whether  the  voltage  is  given 
at  the  load  or  source.  The  value  of  M  is  then  calculated  from 
formula  (46)  or  (47),  and  the  size  of  wire  is  noted  from  Tables  18 
to  27,  opposite  the  required  value  of  M. 

When  given  the  per  cent  power  loss,  R  is  calculated  from  for- 
mula (48)  or  (49),  and  the  size  of  wire  is  found  in  Table  3,  page  7, 
opposite  the  required  resistance  per  mile.  The  wire  found  by 
either  method  is  the  required  size  of  each  conductor. 

40.  Per  Cent  Volt  Loss.  —  It   is    desirable    to    calculate   the 
per  cent  volt  loss  after  the  size  of  wire  is  determined.     The  value 
of  M  whether  between  the  section  letters  a  and  b,  b  and  c,  or  c  and 
d,  is  noted  from  the  proper  table.     The  value  of  V"  or  F0"  is 
calculated  from  formula  (50)  or  (51),  and  then  located  in  Table 
15,  page  51,  in  the  column  headed  by  the  same  section  letters  as 
noted  above  for  M.     The  corresponding  value  of  per  cent  volt 
loss  in  terms  of  the  volts  at  load  or  source,  respectively,  is  then 
obtained  in  the  column  headed  by  V  or  V0.     Where  the  calcu- 
lated value  does  not  appear  in  the  table,  the  corresponding  value 
of  V  or  V0  is  easily  found  by  interpolation. 

The  remaining  calculations  are  clearly  outlined  in  the  table  of 
formulas  on  page  50.  When  the  voltages  at  the  load  and  source 
become  known,  either  formula  for  any  required  item  may  be  used. 


38  TRANSMISSION  CALCULATIONS 

41.   Charging   Current.  —  The   total   charging   current  for   a 
single-phase  circuit  or  for  each  phase  of  a  two-phase  four-wire 

circuit  is  0.000122  EJL 

2  D 
log,,  -j- 

The  charging  current  per  wire  of  a  three-phase  three-wire 
circuit  is  0.000141  EJL 


E0  =  Kilovolts  between  wires  at  source. 

/  =  Frequency  in  cycles  per  second. 

L  =  Distance  from  source  to  load,  in  miles. 

D  =  Distance  between  center  of  wires. 

d  =  Diameter  of  wire  in  same  unit  as  D. 
42.  Capacity  Effects.  —  Capacity  influences  the  voltage  loss, 
the  power  loss  and  the  power-factor  at  generator.  In  all  systems 
except  those  of  unusual  length  or  high  commercial  frequency, 
capacity  may  be  entirely  neglected  without  any  disturbing  error. 
However,  the  effect  of  capacity  may  be  easily  included  in  the 
calculations  by  assuming  that  the  same  result  is  produced  by 
substituting  for  the  distributed  capacity  of  the  line  one-half  its 
total  at  each  end.*  The  true  power-factor  of  the  load  is  then 
replaced  by  an  apparent  value  determined  as  follows: 

.     .'....       (44) 

After  if  has  been  calculated,  the  corresponding  apparent 
power-factor  is  taken  from  Table  14,  page  39-,  and  this  value  is 
used  in  the  table  of  M  for  the  determination  of  the  size  of  wire, 
volt  loss  and  power  loss. 

The  true  power-factor  at  the  source  is  obtained  by  the  follow- 
ing method:  Formula  (61)  is  solved  for  K0'  ',  using  the  apparent 
power-factor  of  load  for  K.  The  value  of  t0'  corresponding  to 
KQ  is  noted  from  Table  14,  page  39,  and  substituted  in  formula 
(45)  for  the  reactance  factor.  The  true  power-factor  at  source 
is  then  obtained  from  Table  14,  by  finding  K0  corresponding  to 
t0.  (See  Example  20,  page  48,  and  Example  25,  page  69.) 


(45) 

WQ 

*  Method  of  H.  Fender,  Proceedings  of  A.I.E.E.,  June,  1908,  page  772. 


ALTERNATING-CURRENT   TRANSMISSION 


39 


In  formulas  (44)  and  (45)  above, 

c  =  Capacity  factor.     Table  13,  page  39. 

E  =  Kilovolts  between  wires  at  load. 

EQ  =s  Kilovolts  between  wires  at  source. 

/    =  Frequency  in  cycles  .per  second. 

L  =   Distance  from  source  to  load,  in  miles. 

=  Reactance  factor  for  true  power-factor  of  load.  Table  14. 
=  Reactance    factor   for    apparent    power   factor   of   load. 
=  Reactance  factor  for  true  power-factor  at  source. 
=  Reactance  factor  for  apparent  power-factor   at  source. 
W  =  Kilowatts  at  load. 
W0=  Kilowatts  at  source. 

Table  13.  — Values  of  c  for  Overhead  Wires. 


Cir.  Mils  or  A.W.G. 

1-Phase  System,  or 
2-  Phase  4-Wire  System 

3-Phase  3-Wire 
System. 

Distance  Between  Center  of  Wires,  in  Inches. 

18 

36 

60 

18 

36 

60 

1,000,000  to  500,  000 
500,000  to  0000 
0000  to  8 

0.00004 
0.00004 
0.00003 

0.00003 
0.00003 
0.00003 

0.00003 
0.00003 
0.00002 

0.00008 
0.00007 
0.00006 

0.00007 
0.00006 
0.00005 

0.00006 
0.00006 
0.00005 

Table  14.  —  Reactance  Factors. 


Power-factors  in  per  cent. 

95 

Lead. 

98 

Lead. 

100 

98 

Lag. 

95 

Lag. 

90 

Lag. 

85 

Lag. 

80 

Lag 

75 
Lag. 

Reactance  factors  

-0.33 

-0.20 

0 

0.20 

0.33 

0.49 

0.62 

0.75 

0.88 

43.  Examples.  —  Examples  illustrating  the  application  of  the 
formulas  on  page  50  have  been  chosen  at  random  and  are  solved 
below  in  detail.  The  directness,  simplicity,  range  and  accuracy 
of  the  method  will  at  once  be  apparent.  As  a  comparison  some 
of  the  examples  are  based  on  problems  already  in  print,  although 
in  most  cases  direct  solutions  are  impossible  by  other  methods. 
In  order  to  prove  the  accuracy  of  the  method,  the  exact  result 


40  TRANSMISSION  CALCULATIONS 

has  been  inclosed  in  brackets   after   many  of    the    calculated 
values. 

Example  n.  —  A  transformer  substation  of  an  electric  railway 
is  to  be  supplied  with  3200  kw.  at  15  cycles  per  second  and  80 
per  cent  lagging  power-factor  over  40  miles  of  single-phase  copper 
line  with  its  wires  on  60-inch  centers.  The  loss  is  to  be  15  per 
cent  of  the  33,000  volts  generated,  the  temperature  20°  cent.,  and 
the  wires  are  to  be  of  100  per  cent  conductivity. 

GIVEN  ITEMS. 

W  =  3200  kw.;       EQ  =  33  kv.;       70  =  15  per  cent; 
L  =  40  miles;     K  =  0.80  lag. 

From  Table  16,  page  52,     A  =  0.200. 
From  Table  15,  page  51,     V0'  =  10.9. 

REQUIRED  ITEMS. 

Size  of  each  wire.  —  From  (47), 

M=      (33)2X10.9 
0  .  2  X  3200  X  40 

From  Table  18,  page  53,  use  No.  00,  for  which  M  =  0.494 
(o  -  d). 

Per  cent  volt  loss.  —  From  (51), 

T/  „  __  0.494  X  0.2  X  3200  X  40  =  n  6  fc_^\ 

From  Table  15,  page  51, 

V0=  15.3  per  cent.  [15.3  per  cent.] 

Kttovolts  at  load.  —  From  (53), 

E  =  33  (1  -  0.153)  =  28.0  kv.  [28.0  kv.] 

VoUloss.  —  From  (54), 

v  =  1000  (33  -  28)  -  5000  volts.  [5000  volts.] 
Per  cent  power  loss.  —  From  Table  3,  page  7, 

R  =  0.411  ohm. 


[21.0  per  cent.] 
Kilowatts  at  source.  —  From  (57), 

W0=  3200  (1  +  0.21)  =  3870  kw.  [3870  kw.] 


ALTERNATING-CURRENT   TRANSMISSION  41 

Watt  loss.  —  From  (58), 

P=  1000  (3870-3200)  =  670,000  watts.    [670,000  watts.] 
Amperes  per  wire.  —  From  Table  17,  page  52,  B  =  1.000. 
From  (60), 


33  (1  - 

From  Table  4,  page  8,  No.  00  conductors  will  safely  carry  this 
current. 

Power-factor  at  source.  —  From  (61), 

£0=  (1  +  0.21)  (1  -  0.153)  0.80  =  0.820.        [0.820.] 

Example  12.  —  A  lighting  transformer  10,000  feet  distant  is 
to  receive  50  kw.  at  95  per  cent  lagging  power-factor  and  125 
cycles  per  second  with  a  loss  of  5  per  cent  of  the  2000  volts  at 
load  at  a  temperature  of  30°  cent.  The  line  is  to  be  single-phase 
and  of  copper  conductors  of  100  per  cent  conductivity  on  18- 
inch  centers. 

GIVEN   ITEMS. 

T7=50kw.;       #=2kv.;  7=  5  per  cent; 

L  =  10,000  feet;     K  =  0.95  lag. 

From  Table  16,  page  52,     A  =  0.0393. 
From  Table  15,  page  51,     V  =  4.8. 

REQUIRED  ITEMS. 

Size  of  each  wire.  —  From  (46), 

^  (2)2  X  4.8 

0.0393  X  50  X  10 

From  Table  22,  use  No.  0,  for  which  M  =  0.905  (b  -  c). 
Per  cent  volt  loss.  —  From  (50), 

v/,  =  0.905  X  0.0393  X  50  X  10  =  445/5         \ 

From  Table  15,  page  51,  7=  4.75  per  cent.    [4.75  per  cent.] 
Kilovolts  at  source.  —  From  (52), 

E0=  2  (1  +  0.0475)  =  2.10  kv.  [2.10  kv.] 

Per  cent  power  loss.  —  From  Table  3,  page  7,  R  =  0.518  ohm. 


42  TRANSMISSION  CALCULATIONS 

From  (55), 


Kilowatts  at  source.  —  From  (57), 

W0=  50  (1  +  0.0282)  =  51.4  kw.          [51.4  kw.] 
Amperes  per  wire.  —  From  Table  17,  page  52,  A  =  1.000. 

From  (59),        /  =  —  ^_  =  26.3  amperes.      [26.3  amperes.] 
2  X  0.95 

Power-factor  at  source.  —  From  (61), 


Example  13.  —  (From  Electric  Journal,  1907,  page  231,  Ex.  2.) 

GIVEN  ITEMS. 
TF=75kw.;         #=2kv.;         Wires  =  No.  4; 

L  =  5000  feet;        K  =  0.95  lag. 

System:  1-phase,  60  cycles,  18-inch  spacing. 
Assume  wires  of  100  per  cent  copper  and  temperature  of  20° 
cent. 

From  Table  16,  page  52,  A  =  0.0379. 
From  Table  21,  page  56,  M=  1.38  (c  -  d). 

REQUIRED  ITEMS. 

Per  cent  volt  loss.  —  From  (50), 

r/,  _  1.38X0.0379  X  75  X  5_  4  9(Hc  _  ^ 

(2)2 

From  Table  15,  page  51, 

V  =  5.40  per  cent.  [5.40  per  cent.] 

Kilovolts  at  source.  —  From  (52), 

E=  2  (1  +  0.054)  =  2.11  kv.  [2.11  kv.] 

Volt  loss.  —  From  (54), 

v=  1000  (2.11  -  2.00)  =  110  volts.      [110  volts.] 

Example  14.  —  (From  "Alternating  Currents  "  by  Franklin  and 
Esty,  1907,  page  321.) 


ALTERNATING-CURRENT  TRANSMISSION  43 

GIVEN  ITEMS. 

TF=1000kw.;     #0=23kv.;         E  =  20  kv.; 
L  =  30  miles;     K  =  0.85  lag. 

System:  1-phase,  60  cycles,  18-inch  spacing. 
Assume  temperature  of  20°  cent,  and  copper  of  100  per  cent 
conductivity. 

From  Table  16,  page  52,  A  =  0.200. 

7=  -^—=15  per  cent.     From  Table  15,  page  51,  V  =  13.1. 

REQUIRED  ITEMS 

Size  of  each  wire.  —  From  (46), 
(20)2X13.1 


0.2  X  1000  X  30 

From  Table  21,  page  56,  use  No.  1,  for  which 
M=  0.936  (c  -  d}. 

Per  cent  volt  loss.  —  From  (50), 

T/"  _  0.936  X  0.2  X  1000  X  30  __+Anff       ^ 

~7207" 

From  Table  15,  page  51,  V  =  15.75  per  cent.  [15.5  per  cent.] 
Example  15.  —  A  load  of  750  kw.  of  2-phase  power  at  25  cycles 
per  second  and  100  per  cent  power-factor  is  to  be  delivered  over 
5  miles  of  aluminum  line  of  4  wires  with  36  inches  between  con- 
ductors of  the  same  phase,  with  a  loss  of  7.5  per  cent  of  the  6600 
volts  generated,  at  30°  cent. 

GIVEN   ITEMS. 

W=  750  kw.;       E0=  6.6  kv.;     70=  7.5  per  cent; 
L  =  5  miles;      K=  1.00. 

From  Table  16,  page  52,  A    =  0.104. 
From  Table  15,  page  51,  F0'=  6.4. 

REQUIRED  ITEMS. 

Size  of  each  wire.  —  From  (47), 

M-     («•«)'  xa.4  --  0716. 

0.104  X  750  X  5 


44  TRANSMISSION   CALCULATIONS 

From  Table  23,  page  58,  use  No.  0,  for  which 

M  =  0.755  (c  -  d). 
Per  cent  volt  loss.  —  From  (51), 

y  „     0.755X0.104X75X5  =  6  ?6  (c  _  ^ 
(6.6)2 

From  Table  15,  page  51, 

F0=  8.16  per  cent.  [8.10  per  cent.] 

Kilovolts  at  load.  —  From  (53), 

E  =  6.6  (1  -  0.0816)  =  6.06  kv.  [6.07  kv.] 

Per  cent  power  loss.  —  From  Table  3,  page  7,  R  =  0.837  ohm. 
From  (55), 

p  =  0.837  X  0.104  X  750  X  5  =88Qpercent      [8.82  per  cent>] 
(6.06) 

Kilowatts  at  source.  —  From  (57), 

W  =  750  (1  +  0.0889)  =  817  kw.  [816  kw.] 

Amperes  per  wire.  —  From  Table  17,  page  52,  B  =  0.500. 

From  (59),        7  =  °'5  X  75Q  =  61.9  amperes.     [61.8  amperes.] 
6.06 

From  Table  4,  page  8,  No.  0  aluminum  conductors  will  easily 
carry  this  current. 

Power-factor  at  source.  —  From  (61), 

K0=  (1  +  0.0889). (1  -  0.0816)  =  1.00.  [1.00.] 

Example  16.  —  A  two-phase  three-wire  line  30  miles  long  is 
to  deliver  2000  kw.  at  40  cycles  per  second  and  98  per  cent  lag- 
ging power-factor  with  a  loss  of  10  per  cent  of  the  20,000  volts  at 
load.  Calculate  the  size  of  copper  wires  of  98  per  cent  conduc- 
tivity on  36-inch  centers  for  a  temperature  of  50°  cent. 

GIVEN   ITEMS. 

W=  2000  kw.;       E=  20  kv.;          V=  10  per  cent; 
L=  30  miles;      K  =  0.98  lag. 

From  Table  16,  page  52, 'A  =  ^~  =  0.138. 

U.  C/O 

From  Table  15,  page  51,  V  =  9.1. 


ALTERNATING-CURRENT   TRANSMISSION  45 

REQUIRED  ITEMS. 

Size  of  each  wire.  —  From  (46), 

M= 

0.138  X  2000  X  30 

From  Table  20,  page  55,  use  three  No.  00  wires,  or  increase 
the  common  lead  to  No.  000.     For  No.  00,   M  =  0.466     (c  -  d). 
Per  cent  volt  loss.  —  From  (50), 

T///  _  0.466  X  0.138  X  2000  X  30  _  Q  R4  /         ^ 

(20)2 

From  Table  15,  page  51,  7=  10.84  per  cent. 
Since  the  volt  loss  in  the  common  lead  is  1.41  times  that  in  each 
of  the  outer  wires, 

Per  cent  volt  loss  in  outer  wire  =  -  :  -  =  4.50  per  cent. 

With  a  common  lead  one  number  larger  in  the  A.  W.  Gauge  (26 
per  cent  larger)  than  each  outer  wire, 

Approximate  per  cent  volt  loss  in  common  lead 

=  10.84-  4.5 
1.26 

Therefore,  in  the  above  case  with  No.  00  outer  wires  and  No. 
000  for  the  common  lead, 

Approximate  total  per  cent  volt  loss  =  4.50  +  5.03  =  9.53  per 
cent. 

Amperes  per  wire.  —  From  Table  17,  page  52,  B  =  0.500. 

From  (59),  in  each  outer  wire, 


In  common  wire,      /=  51  X  1.41  =  71.9  amperes. 

Example  17.  —  Two  substations  located  respectively  at  40 
and  50  miles  away  are  supplied  from  a  No.  00,  25-cycle  three- 
phase  aluminum  line  with  wires  on  60-inch  centers.  The  total 
load  at  the  nearer  substation  (No.  1)  is  2000  kw.  at  95  per  cent 
lagging  power-factor,  while  the  other  (No.  2)  takes  1000  kw.  at 
98  per  cent  lagging  power-factor,  the  generator  voltage  being 
33,000  and  the  temperature  30°  cent. 


46  TRANSMISSION  CALCULATIONS 

GIVEN  ITEMS. 

For  No.  1 :      W=  2000  kw. ;    E0  =  33  kv. ;    L  =  40  miles ; 

K  =  0.95  lag;      M  =  0.690  (c  -  d) . 
For  No.  2:      W=  1000  kw.;     L  =  10  miles;    K  =  0.98  lag; 

M=  0.655  (c  -  d). 
From  Table  16,  page  52,  A  =  0.104. 

REQUIRED  ITEMS. 

Per  cent  volt  drop  to  No.  1.  —  From  (51), 

T/  //  _  0.690  X  0.104  X  2000  X  40  _  -  9ft  .          A 
VQ    :  (33)2 

From  Table  15,  page  51,  V0=  6.26  per  cent. 

KilovoltsatNo.  1.—  From  (53),  #  =  33  (1  -  0.0626)  =  30.9  kv. 
Per  cent  volt  drop  from  No.  1  to  No.  2.  —  E0  =30.9  kv.  from 
above. 

From  (51), 

T/  „  _  0.655  X  0.104  X  1000  X  10  _  n  71  /        ^ 
VQ    :  (30.9)2 

From  Table  15,  page  51,  V0=  0.81  per  cent. 

Kilovolts  at  No.  2.— From  (53),  E  =  30.9  (1  -0.0081)  =  30.6  kv. 

Example  18.*  —  A  three-phase  load  of  5000  kw.  and  25  cycles 
is  to  be  delivered  at  30,000  volts  to  a  receiver  having  95  per  cent 
lagging  power-factor,  over  40  miles  of  No.  00  copper  wires  of 
100  per  cent  conductivity  on  48-inch  centers,  at  a  temperature 
of  20°  cent. 

GIVEN  ITEMS. 

W=  5000  kw.;    E=  30  kv.;   L=  40  miles;    K=  0.95  lag. 
By  interpolation  from  Table  19,  page  54,  M  =  0.457  (c  —  d). 
From  Table  16,  page  52,  A  =  0.100. 

REQUIRED  ITEMS. 

Per  cent  volt  loss.  —  From  (50), 

T///  _  0.457  X  0.1  X  5000  X  40  _  n  A  n  ,         ,. 
""  }* 


Problem  of  H.  Fender,  Proceedings  of  A.I.E.E.,  June,  1908,  page  771. 


ALTERNATING-CURRENT   TRANSMISSION  47 

From  Table  15,  page  51,  V  =  11.3  per  cent.    [11.3  per  cent.] 
Kilovolts  at  source.  —  From  (52), 

EQ=  30  (1  +  0.113)  =  33.4  kv.  [33.4  kv.] 

Volt  loss.  —  From  (54), 

v  =  1000  (33.4  -  30.0)  =  3400  volts.          [3400  volts.] 

Per  cent  power  loss.  —  From  Table  3,  page  7,  R  =  0.411  ohm. 
From  (55), 


Kilowatts  at  source.  —  From  (57), 

WQ=  5000  (1  +  0.101)  =  5505  kw.  [5505  kw.] 

Amperes  per  wire.  —  From  Table  17,  page  52,  B  =  0.578. 
From  (59), 

7  =  °'578  X  500°  =  101.4  amperes.      [101.4  amperes.] 

From   Table   4,  page  8,   No.  00  wires   have   ample   current- 
carrying  capacity. 

Power-factor  at  source.  —  From  (61), 

=  o.94  lag.  [0.94  lag.] 


Example  19.  —  A  three-phase  load  of  10,000  kw.  at  60  cycles 
per  second  and  98  per  cent  leading  power-factor  is  to  be  delivered 
over  65  miles  of  three  aluminum  wires  on  60-inch  centers,  with 
a  loss  of  15  per  cent  of  the  44,000  volts  generated;  temperature 
40°  cent. 

GIVEN  ITEMS. 

W  =  10,000  kw.  ;     E0  =  44  kv.  ;      F0  =  15  per  cent; 
L  =  65  miles;  K=  0.98  lead. 

From  Table  16,  page  52,     A  =  0.108. 
From  Table  15,  page  51,     TV  =  10.9. 

REQUIRED  ITEMS. 

.    Size  of  each  wire.  —  From  (47), 

M==         (44)2  X  10.9 

0.108  X  10,000  X  65 


48  TRANSMISSION  CALCULATIONS 

From  Table  26,  page  61,  use  No.  0000,  for  which  M  =  0.316 
(a  -  6). 

Per  cent  volt  loss.  —  From  (51), 

T,  „  _  0.316  X  0.108  X  10,000  X  65  _  1  1  A  ,         M 
(44)* 

From  Table  15,  page  51,  F0=  17.3  per  cent.  [17.4  per  cent.] 
Per  cent  power  loss.  —  From  Table  3,  page  7,  R  =  0.417  ohm. 
From  (56), 


Power-factor  at  source.  —  From  (61), 

K0=  (1  +  0.23)  (1  -  0.173)  0.98=  1.00.      [0.994  lead.] 

Example  20.*  —  A  three-phase  load  of  7500  kw.  at  90  per 
cent  lagging  power-factor  and  60  cycles  per  second  is  to  be 
delivered  over  140  miles  of  three  copper  wires  of  100  per  cent 
conductivity  on  96-inch  centers,  with  a  voltage  loss  of  18.7  per 
cent  of  the  volts  at  load,  the  voltage  at  the  source  being  71,200. 
Include  the  effect  of  capacity  and  assume  a  temperature  of 
20°  cent. 

GIVEN   ITEMS. 

W=  7500  kw.;     £J0=  71.2  kv.;      V=  18.7  per  cent; 
L=  140  miles;    K=  0.90  lag. 

From  Table  16,  page  52,  A  -  0.100. 

From  (63),     V0  =  —  |yj^  -  15-8  per  cent. 

From  Table  15,  page  51,  T  V  =  11.2. 

REQUIRED  ITEMS. 

Apparent  power-factor  of  load.  —  Assuming  that  the  required 
size  of  wire  will  be  between  Nos.  0000  and  8,  c  is  approxi- 
mately 0.00005.  (From  Table  13,  page  39.) 

From  Table  14,  page  39,  the  reactance  factor  corresponding 
to  90  per  cent  lagging  power-factor  =  0.49. 

*  Based  on  Problem  of  H.  Fender,  Proceedings  of  A.I.E.E.,  June,  1908, 
page  774. 


ALTERNATING-CURRENT  TRANSMISSION  49 

From  (44),  page  38, 

,  =  Q  4Q  _  Q.QOQQ5[71.2  (1  -  0.158)]2  60  X  140  =  Q  2g 

7500 

By  interpolation  in  Table  14,  page  39,  the  apparent  power- 
factor  of  load,  K',  =  0.96  lag 
Size  of  each  wire.  —  From  (47), 


0.1  X  7500  X  140 

From  Table  21,  page  56,  use  No.  00,  for  which  M  is  found  by 
interpolation  to  be  0.600  (c  —  d). 
Per  cent  volt  loss.  —  From  (51), 

T7  „      0.600  X  0.1  X  7500  X  140  _  10  A  (          ,, 
T°    :  (71.2)' 

From  Table  15,  Page  51,F0  =  16.7  per  cent.    [17.2  per  cent.] 

Kilovolts  at  load.  —  From  (53), 

E  =  71.2  (1  -  0.167)  =  59.3  kv.  [59.0  kv.] 

Per  cent  power  loss.  —  From  Table  3,  page  7,  R  =  0.411  ohm. 
From  (55), 


Kilowatts  at  source.  — From  (57), 

W0  =  7500  (1  +  0.133)  =  8500  kw.          [8500  kw.] 
Power-factor  at  generator  (not  including  capacity).  —  From  (61), 
K0'=  (1  +  0.133)  (1  -  0.167)  0.96  =  0.91. 

From  Table  14,  page  39,  ?0'  m  0.46. 
From  (45),  page  38, 

0.00005  X(71.2)2  X  60  X  140      n  91 
8500 

From  Table  14,  page  39,  the  true  power-factor  at  load,  K0 
(corresponding  to  Z0),  =  0.98  lag.  [0.98  lag.] 


50 


TRANSMISSION  CALCULATIONS 


Formulas  for  A.  C.  Transmission  by  Overhead  Wires. 


Required  Items. 

For  Voltage  Given  at  Load. 

For  Voltage  Given  at  Source. 

Size  of  each  wire 
from 
per  cent  volt  loss. 

A  from  Table  16,  page  52. 

V  from  Table  15,  page  51. 

VQ'  from  Table  15,  page  51. 

M-E2V'             (16) 

Jf-Sffii'                  • 
AWL 

AWL   ••;•<« 

Then  find  wire  from  Tables  18  to  27,  pages  53  to  62. 

Size  of  each  wire 
from 
per  cent  power  loss. 

„      PE2K2 

E_P[E,(l-OMVn)KV 

R  —  •              .          (48) 
AWL 

AWL 

(49) 

Then  find  wire  from  Table  3,  page  7. 

Per  cent  volt  loss. 

V"             MAWL                       ,rftx 

T///.     MAWL                - 

J£2        •     '     •  (&UJ 

T°                ^ 

V  from  Table  15,  page  51. 

FQ  from  Table  15,  page  51. 

Kilovolts  at  source 
or  load. 

E0  =  E  (1  +  0.  OIF).  (52) 

E  =  #0(1-0.01F0).  (53) 

Volt  loss. 

v  =  1000  (E0  -  E)  =  10  VE  =  10  V0E0  .   .    .  (54) 

Per  cent  power  loss. 

R  from  Table  3,  page  7. 

p      RA  WL 

p_            RAWL 

FJT  H      0  01  V  \'K"\* 

E*K*    ' 

lJ2jO\i       U.U1K0JJVJ3 

(56) 

Kilowatts  at  source. 

W,  =  w  (H 

-  0.01P)     ,                   .  (57) 

Watt  loss. 

j>=  1000(JF0-TF)  =  10  PW.    .  (58) 

Amperes  per  wire. 
(See  par.  34,  page  35.) 

B  from  Table  17,  page  52. 

2  _  BW                     ™ 

I-            BW            (60) 

EK     '    '    '    '  (     ' 

^0(l-0.01F0)^'k 

Power-factor  at  source. 

K       (l+0.01P)JK_rl 
K°  ~     1  +  0.017 

o  01  PUI    001F^7^  (&\\ 

When  given  E  and  V0 
When  given  J8Q  and  V 
When  given  W,  find 

find  F  from  F-  F0/(l  - 

find  F0froinF0  =  F/(l 
W  from  TF  -  TF'JT 

-  0.01  F0)  ...       .      (62) 

+  0.01F)  (63) 

.      (64r\ 

NOTATION. 


A  =  Transmission  factor.  Table  16,  page  52. 
B  =  Transmission  constant.  Table  17,  p.  52. 
E  =  Kilovolts  (1000  volts)  between  wires  at 

load. 
E0=  Kilovolts  (1000  volts)  between  wires  at 

source.    See  par.  32,  p.  35. 
/  =  Amperes  per  wire.     See  par.  32,  p.  35. 
K  —  Power-factor  of  load,  in  decimals. 
K0=  Power-factor  at  source,  in  decimals. 
L  =  Distance  from  source  to  load,  in  miles 

or  1000  feet. 

M  =  Wire  factor  Tables  18  to  27,  pp.  53  to  62. 
P  =  Power  loss  in  per  cent  of  power  at  load. 


»    = 

H  = 


Power  loss  in  watts. 

Resistance  per  mile  of   one  wire,  in 

ohms.    Table  3,  p.  7. 
=  Volt  loss  in  per  cent  of  volts  at  load. 
=  Volt  loss  in  per  cent  of  volts  at  source. 
*Y,  F",  F0"=  Volt  loss  factors.    Table 

15,  p.  51 
=  Volt  loss  (volts  at  source  —  volts  at 

load). 

=  Real  kilowatts  at  load. 
=  Real  kilowatts  at  source. 
=  Kilovolt-amperes  (apparent  power)  at 
load. 


ALTERNATING-CURRENT  TRANSMISSION 


51 


Table  15.  —  Values  of  Volt  Loss  Factors. 


V 

V" 

V0" 

y 

V" 

V0" 

or 

V 

Vo 

or 

V 

V0' 

Vo 

a-b 

b-c 

c-d 

a-6 

b-c 

c-d 

V0 

a-6 

b-c 

c~d 

a-6 

b-c 

c-d 

On 

00 

On 

0  0 

0  0 

0  0 

0  0 

0  0 

0  0 

8  0 

7  4 

.  U 

0.2 

.u 
0.2 

.V 

0.2 

0.2 

0.2 

0.2 

0.2 

0.2 

0.2 

8.2 

7.6 

7.7 

7.6 

7.3 

6.9 

7.0 

6.9 

6.8 

0.4 

0.4 

0.4 

0.4 

0.4 

0.4 

0.4 

0.4 

0.4 

8.4 

7.8 

7.9 

7.8 

7.5 

7.0 

7.1 

7.1 

6.9 

0.6 

0.6 

0.6 

0.6 

0.5 

0.6 

0.6 

0.6 

0.6 

8.6 

7.9 

8.0 

8.0 

7.7 

7.2 

7.3 

7.2 

7.1 

0.8 

0.8 

0.8 

0.8 

0.7 

0.8 

0.8 

0.8 

0.7 

8.8 

8.1 

8.2 

8.1 

7.9 

7.3 

7.4 

7.4 

7.2 

.0 

.0 

.0 

.0 

0.9 

.0 

.0 

.0 

0.9 

9.0 

8.3 

8.4 

8.3 

8.0 

7.5 

7.5 

7.5 

7.4 

.2 

.2 

.2 

.2 

1.1 

If 

.2 

.2 

.2 

.1 

9.2 

8.4 

8.5 

8.5 

8.2 

7.6 

7.6 

7.6 

7.5 

7« 

.4 
.6 

.4 
.6 

.4 
.6 

.4 
.5 

.  J 

1.5 

.6 

.6 

.5 

.4 

9.6 

8.8 

8.9 

8.8 

8.5 

7.8 

7.9 

7.9 

,  j 

7.8 

.8 

.8 

.8 

.7 

1.7 

.7 

.8 

.7 

.6 

9.£ 

8.9 

9.0 

9.0 

8.7 

8.0 

8.0 

8.1 

7.9 

2.0 

2.0 

2.0 

1.9 

1.8 

1.9 

2.0 

1.9 

.8 

10.0 

9.1 

9.2 

9.1 

8.9 

8.1 

8.1 

8.2 

8.1 

2.2 

2.2 

2.2 

2.1 

2.0 

2.1 

2.2 

2.1 

.9 

10.2 

9.2 

9.3 

9.3 

9.1 

8.2 

8.2 

8.3 

8.2 

2.4 

2.3 

2.4 

2.3 

2.2 

2.3 

2.3 

2.2 

2.1 

10.4 

9.4 

9.5 

9.5 

9.3 

8.4 

8.4 

8.5 

8.4 

2.6 

2.5 

2.6 

2.5 

2.4 

2.5 

2.5 

2.4 

2.2 

10.6 

9.6 

9.7 

9.7 

9.5 

8.5 

8.5 

8.6 

8.5 

2.8 

2.7 

2.8 

2.7 

2.6 

2.t 

2.7 

2.6 

2.4 

10.8 

9.7 

9.8 

9.8 

9.6 

8.6 

8.6 

8.8 

8.7 

3.0 

2.9 

3.0 

3") 

2.8 

3/\ 

2.7 

2/\ 

2.8 

3/t 

2.9 

2.8 

3r\ 

2.6 

20 

11.0 

9.9 

1  A    1 

10.0 

1  f\     1 

10.0 

ml 

9.8 

8.7 

8.7 

8.9 

8.9 

3.2 
3.4 

3.  1 
3.3 

»2 

3.4 

.0 

3.2 

.V 

3.1 

.U 

3.2 

3.  1 

3.2 

.U 

3.1 

.O 

2.9 

11.2 
-11.4 

Iv.  I 

10.2 

IU.  1 

10.3 

.  1 

10.3 

10.2 

9.0 

8.9 

9.1 

9.1 

3.6 

3.5 

3.6 

3.4 

3.3 

3.3 

3.4 

3.3 

3.1 

3     a 

11.6 

10.4 

|  A     7 

10.4 
in  ft 

10.4 

in    o 

10.3 

in    n 

9.1 

9-7 

9.0 

9' 

9.3 

9C 

9.3 

9e 

3.8 
4.0 

3.8 

4.0 

3.8 

3.6 

3.7 

3.8 

3.6 

.J 

3.4 

12.  C 
12.5 

IU.  / 

11.1 

IU  .0 

ii.  i 

IU  .C 

11.3 

IU.  / 

11.1 

.  J 

9.6 

./ 
9.5 

9.8 

9.8 

4.2 

4.0 

4.2 

4.0 

3.8 

3.8 

3.9 

3.8 

3.6 

13.0 

11.5 

11.5 

11.7 

11.6 

9.8 

9.7 

10.1 

10.2 

4.4 

4.2 

4.4 

4.2 

4.0 

4.0 

4.1 

4.0 

3.8 

13.5 

11.9 

11.9 

12.1 

12.0 

10.1 

9.9 

10.4 

10.5 

4.6 

4.4 

4.5 

4.3 

4.2 

4.2 

4.3 

4.1 

3.9 

14.0 

12.3 

12.2 

12.6 

12.5 

10.4 

10.2 

10.7 

10.8 

4.8 

4.6 

4.7 

4.5 

4.4 

4.3 

4.4 

4  1 

4  1 

14  5 

1?  7 

1?  6 

13  0 

1?  9 

10  6 

10  4 

11  (I 

11  1 

5.0 

4.8 

4.9 

4.7 

4.5 

4.5 

4  6 

4  4 

4  3 

15  0 

13  1 

n  o 

n  4 

13  4 

10  9 

10  6 

11  ? 

11  4 

5.2 

4.9 

5.1 

4.9 

4./ 

4.7 

4.8 

4.6 

4.4 

15.5 

13.4 

13.3 

13.8 

13.8 

11.1 

10.8 

11.5 

11.7 

5.4 

5.1 

5.3 

5.1 

4.9 

4.8 

4.9 

4.8 

4.6 

16.0 

13.8 

13.6 

14.2 

14.2 

11.3 

11.0 

11.8 

12.0 

5.6 

5.3 

5.5 

5.3 

5.1 

5.0 

5.1 

4.9 

4.7 

16.5 

14.2 

14.0 

14.6 

14.7 

11.5 

11.2 

12.0 

12.3 

5.8 

5.5 

5.7 

5.5 

5.2 

5.1 

5.3 

5.1 

4.9 

17.0 

14.5 

14.3 

15.0 

15.1 

11.7 

11.3 

12.3 

12.6 

6.0 

5.7 

5.8 

5.6 

5.4 

5  1 

5  4 

5  3 

5  1 

17  5 

14  9 

14  7 

15  4 

15  6 

11  9 

II  5 

1?  5 

12.9 

6.2 

5.8 

6.0 

5.8 

5.6 

5.4 

5.6 

5.4 

5.2 

18.0 

15.3 

15.0 

15.8 

16.0 

12.1 

11.6 

12.7 

13.2 

6.4 

6.0 

6.2 

6.0 

5.8 

5.6 

5.7 

5.6 

5.4 

18.5 

15.6 

15.3 

16.2 

16.5 

12.3 

11.8 

13.0 

13.4 

6.6 

6.2 

6.4 

6.2 

5.9 

5.8 

5.9 

5.7 

5.5 

19.0 

16.0 

15.6 

16.6 

16.9 

12.5 

11.9 

13.2 

13.7 

6.8 

6.4 

6.5 

6.3 

6.1 

5.9 

6.0 

5.9 

5.7 

20.0 

16.7 

16.2 

17.4 

17.8 

12.8 

12.2 

13.6 

14.2 

7.0 

6.5 

6.7 

6.5 

6.3 

6.1 

6.2 

6.0 

5.8 

21.0 

17.4 

16.8 

18.2 

18.7 

7.2 

6.7 

6.9 

6.7 

6.5 

6.2 

6.3 

6.2 

6.0 

22.0 

18.0 

17.4 

19.0 

19.6 

7.4 

6.9 

7.0 

6.9 

6.6 

6.3 

6.4 

6.3 

6.2 

23.0 

18.7 

18.0 

19.7 

20.5 

.... 

.... 

7.6 

7.1 

7.2 

7.1 

6.8 

6.5 

6.6 

6.5 

6.3 

24.0 

19.4 

18  5 

70  5 

71  4 

7.8 

7.2 

7.4 

7.3 

7.0 

6.6 

6.7 

6.6 

6.5 

25.0 

20.0 

19.1 

21  3 

7?  3 

52 


TRANSMISSION  CALCULATIONS 


Table  16—  Values  of  A  for  Balanced  Loads. 


2-  Phase  Sys- 

Temperature 
In  degrees. 

1-Phase  Sys- 
tem with  2 
Equal  Wires 

2-  Phase  Sys- 
tem with  3 
Equal  Wires. 

tern  with  4 
Equal  Wires, 
or  3-Phase  Sys- 
tem  with   3 

Equal  Wires. 

For  L 

ForL 

ForL 

ForL 

ForL 

For  L 

Cent. 

Fahr 

in 

in 

in 

in 

in 

in 

Miles. 

1000  Ft. 

Miles. 

1000  Ft. 

Miles 

1000  Ft. 

0 

32 

0.185 

0.0351 

0.112 

0,0212 

0  0926 

0.0175 

10 

50 

0.193 

0.0365 

0.116 

0.0220 

0.0963 

0.0182 

20 

68 

0.200 

0.0379 

0.121 

0  0229 

0  100 

0.0190 

30 

86 

0.208 

0.0393 

0.125 

0  0237 

0.104 

0.0197 

40 

104 

0.215 

0.0408 

0.130 

0.0246 

0  108 

0  0204 

50 

122 

0  223 

0.0423 

0.135 

0,0255 

0,112 

0.0212 

The  above  values  are  for  copper  wires  of  100  per  cent  conductivity  and  aluminum  wires  of 
62  per  cent  conductivity,  in  Matthiessen's  standard  scale, 
For  copper  of  other  conductivity,  divide  A  by  given  conductivity 
For  aluminum  of  other  conductivity,  divide  A  by  given  conductivity  and  multiply  by  0.62 


Table  17. —  Values  of  B  for  Balanced  Loads. 


1  -Phase  System. 

2-Phase  System. 

3-  Phase  System. 

1.000 

0.500 

0.578 

ALTERNATING-CURRENT  TRANSMISSION 


53 


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54 


TRANSMISSION  CALCULATIONS 


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ALTERNATING-CURRENT  TRANSMISSION 


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56 


TRANSMISSION  CALCULATIONS 


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ALTERNA  TING-CURRENT   TRANSMISSION 


57 


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CHAPTER   IV. 


ALTERNATING-CURRENT  TRANSMISSION  BY   UNDERGROUND 

CABLES. 

44.  Introduction.  —  The  calculations  for  underground  cables 
are  very  similar  to  those  for  overhead  lines.  The  main  difference 
in  results  is  due  to  the  proximity  of  the  wires,  in  consequence  of 
which  inductance  effects  are  considerably  reduced.  However, 
with  commercial  thickness  of  insulation  such  effects  are  not 
entirely  eliminated  for  the  larger  wires  at  the  higher  frequencies. 
The  difference  may  be  ascertained  by  comparing  the  values  of 
M-  at  100  per  cent  power-factor  in  Tables  34  and  35  with  the 
resistance  per  mile  of  the  same  wires  in  Table  3.  As  the  con- 
ductors in  underground  cables  are  almost  always  copper,  the 
results  have  been  calculated  for  that  metal  alone.  The  calcula- 
tions for  15  cycles  per  second  have  been  omitted. 

The  error  of  the  common  assumption  that  the  resistance  drop 
equals  the  volt  loss  is  indicated  in  Table  28  below  for  insulation 
of  ££  inch  thickness. 

Table  28.  —  Error  in  Per  Cent  of  True  Volt  Loss, 
Assuming  Resistance  Drop  =  Volt  Loss. 


25  Cycles  per  Sec.  II  60  Cycles  per  Sec.l  125  Cycles  per  Sec. 

Size  of 
Wire. 

Lagging  Power  Factor  of  Load  in  Per  Cent. 

A  W  G 

100 

90 

80 

100 

90 

80 

100 

90 

80 

0000 

0 

-1 

+  2 

-2 

-15 

-15 

-8 

-35 

-39 

0 

0 

+  3 

+  10 

0 

_  4 

_    1 

_  3 

-18 

-19 

4 

0 

+  7 

+  15 

0 

+  2 

+   8 

0 

-  5 

-  2 

It  is  obvious  from  the  table  that  the  values  resulting  from  the 
assumption  may  be  much  too  high  or  too  low,  except  near  100 
per  cent  power-factor.  The  error  is  due  to  the  power-factor 
angles  of  the  load  and  line.  (See  Figs.  1  and  2,  page  30.) 

45.   Maximum  Error.  —  The  formulas  apply  for  all  volt  losses 


TRANSMISSION  CALCULATIONS 


less  than  20  per  cent  of  the  generated  volts  or  25  per  cent  of  the 
voltage  at  load.  As  in  the  case  of  overhead  lines  the  formulas 
are  approximate,  but  the  error  in  general  is  much  reduced,  even 
without  subdivision  of  the  wire  factors  into  sections.  The  maxi- 
mum error  in  the  calculation  for  the  per  cent  volt  loss  is  about 
5  per  cent  near  0  per  cent  and  20  per  cent  volt  losses,  but  it 
gradually  reduces  to  a  zero  error  near  10  per  cent  loss.  Thus 
for  a  true  volt  loss  of  5  per  cent  the  extreme  range  of  the  cal- 
culated result  is  about  4.85  to  5.15  per  cent.  Under  ordinary 
conditions  the  error  is  immeasurably  small. 

The  maximum  errors  of  the  formulas  under  the  most  severe 
conditions  are  shown  in  the  following  table. 

Table  29.  —  Maximum  Error  in  Per  Cent  of  True  Values,  at 

20°  Cent. 


For  Voltage 

For  Voltage 

Items. 

Given  at 

Given  at 

Load. 

Source. 

Volt  loss  or  per  cent  volt  loss  

5 

5 

Kilovolts  at  load  or  source  

I 

1 

Power  loss  or  per  cent  power  loss  

0 

2 

Kilowatts  at  source  

0 

0 

Amperes  per  wire  

0 

1 

Power-factor  at  source  

] 

1 

46.  Temperature.  —  The  values  of  M  have  been  calculated  at 
20°  cent,  or  68°  fahr.     The  effect  of  temperature  variations  is 
introduced  in  the  calculations  by  means  of  A,  Table  32,  page  72. 

47.  Properties  of  Conductors.  —  The  values  of  A  are  given  for 
copper  of  100  per  cent  Matthiessen's  standard,  but  any  other 
conductivity  may  be  used  by  dividing  A  of  Table  32  by  the 
given  value  in  the  standard  scale.     It  has  been  assumed  that 
conductors  smaller  than  No.  3  are  solid  and  those  larger  are 
stranded.     However,  any  practical  variation  from  this  assump- 
tion will  have  no  appreciable  effect  on  any  of  the  calculations. 

48.  Thickness  of  Insulation.  —  The  values  of  M  have  been 
calculated  for  a  thickness  of  insulation  of  Jf  inch  on  each  con- 
ductor, but  any  commercial  variation  will  not  materially  affect 
the  results. 

49.  Current-carrying  Capacity.  —  The  current-carrying  capa- 
city in  amperes  per  wire  is  shown  in  Table  4,  page  8,  for  single 


TRANSMISSION  BY   UNDERGROUND  CABLES 


65 


and  three-conductor  cables.  In  most  cases  the  size  of  cable  is 
determined  only  by  the  volt  loss,  but  in  some  problems  the 
current-carrying  capacity  may  be  the  ruling  factor.  Hence  it 
is  advisable  to  note  from  Table  4  that  the  calculated  wire  is 
capable  of  carrying  the  required  current  without  undue  heating. 
50.  Capacity  Effects.  —  Although  underground  lines  are  of 
comparatively  short  length,  yet  the  closeness  of  the  conductors 
and  the  high  values  of  the  specific  inductive  capacity  of  the 
insulation  sometimes  make  it  advisable  to  determine  the  capa- 
city values.  The  method  of  including  capacity  in  the  calcula- 
tion of  volt  loss,  power  loss  and  the  power-factor  at  source  is 
exactly  like  that  given  for  overhead  lines  in  paragraph  42,  page  38. 
The  capacity  factor,  c,  for  underground  cables  is  to  be  taken  from 
Table  30  below.  The  specific  inductive  capacity  was  assumed 
to  be  2.5,  but  for  any  other  value,  c  in  Table  30  should  be  multi- 
plied by  the  ratio  of  the  given  value  to  2.5.  Thus  c  for  a  three- 
phase  cable  having  No.  00  conductors  and  a  specific  inductive 
capacity  of  3,  is 

0.0007X3 


2.5 


=  0.00084. 


Table  30.  —  Values  of  c  for  Underground  Cables, 


Circular  Mils,  or 
A.W.G. 

I-Phase  Cable. 

3-Phase  Cable. 

300,000  to  0 
1  to  6 

0.0003 
0.0003 

0.0007 
0.0006 

For  specific  inductive  capacity  of  insulation  other  than  2.5,  multiply  c  by  given  value 
and  divide  by  2.5. 

Owing  to  upper  harmonics  the  charging  current  in  cables  may 
be  as  much  as  40  per  cent  greater  than  calculated  for  a  sine-wave 
e.  m.  f.  However,  the  effect  cannot  be  included  in  a  general  way 
as  the  increase  depends  entirely  upon  the  generator  design. 

51.  Examples.  —  The  items  to  be  specified  are  the  same  as 
stated  in  paragraph  38,  page  37,  for  overhead  lines.  The  for- 
mulas and  the  method  of  using  them  are  also  similar  except  where 
the  per  cent  volt  loss  is  involved.  In  such  cases  V  is  used  directly 
when  the  voltage  at  load  is  given,  while  F0'"  from  Table  31, 
page  72 ;  is  used  in  formulas  (66)  and  (70)  where  the  voltage  at 


66  TRANSMISSION  CALCULATIONS 

source  is  specified.  The  meaning  of  the  term  "source"  is  specified 
in  paragraph  32,  page  35.  Following  are  typical  examples  illus- 
trating the  application  of  the  formulas  on  page  71  . 

Example  21.  —  A  one-phase  underground  cable  5  miles  long  is 
to  deliver  800  kw.  at  98  per  cent  lagging  power-factor  and  60 
cycles  per  second  with  a  loss  of  10  per  cent  of  the  6600  generated 
volts,  with  copper  of  100  per  cent  conductivity  and  a  temperature 
of  40°  cent. 

GIVEN  ITEMS. 

TF=800kw.;      #0=6.6kv.;    70=  10  per  cent; 
L  =  5  miles;     K  =  0.98  lag. 

From  Table  32,  page  72,  A      =  0.215. 
From  Table  31,  page  72,  1Y"  =  9.0. 

REQUIRED  ITEMS. 

Size  of  each  wire.  —  From  (66), 

M=        (6.6)^X9        =  Q.456. 
0.215  X  800  X  5 

From  Table  35,  page  74,  use  No.  00,  for  which  M  =  0.448. 
Per  cent  volt  loss.  —  From  (70), 

v  ///  -  0.448  X  0.215  X  .8005  _ 
(6.6)2 

From  Table  31,  page  72,  VQ=  9.84  per  cent. 
Kilovolts  at  load.  —  From  (72),  E  =  6.6  (1  -  0.0984)  =  5.95  kv. 
Volt  loss.  —  From  (73),  v=  1000  (6.60  -  5.95)  =  650  volts. 
Per  cent  power  loss.  —  From  Table  3,  page  7,  R  =  0.411  ohm. 
From  (75), 


Kilowatts  at  source.  —  From  (76), 

TF0=  800  (1  +  0.104)  =  883  kw. 

Watt  loss.  —  From  (77),  p  =  1000  (883  -  800)  =  83,000  watts. 
Amperes  per  wire.  —  From  Table  33,  page  72,   B  =  1.000. 
From  (79), 


Power-factor  at  source.  —  From  (80), 

KQ=  (1  +  0.104)  (1  -  0.0984)  0.98=  0.975. 


TRANSMISSION   BY  UNDERGROUND  CABLES  67 

Example  22.  —  A  one-phase  underground  cable  with  No.  1 
copper  conductors  of  98  per  cent  conductivity  and  10,000  feet 
long,  delivers  150  kw.  at  85  per  cent  lagging  power-factor  and 
125  cycles  per  second.  The  generator  voltage  is  2200  and  the 
temperature  is  assumed  50°  cent. 

GIVEN  ITEMS. 

W=  150  kw.;     #0=  2.2  kv.;     L=  10,000  feet;    K=  0.85  lag. 
From  Table  35,  page  74,  M  =  0.896. 

From  Table  32,  page  72,    A  =  Q'0423  =  0.0432. 

\)  «  «yo 

REQUIRED  ITEMS. 

Per  cent  volt  loss.  —  From  (70), 

//,      0.896  X  0.0432  X  150  X  10 


1on 


From  Table  31,  page  72,  V0=  14.0  per  cent. 

Kilovolts  at  load.  —  From  (72),  E  =  2.2  (1  -  0.14)  =  1.89  kv. 

Example  23.  —  A  three-phase  load  of  3000  kw.  and  25  cycles 
per  second  is  to  be  delivered  to  a  receiver  having  a  lagging  power- 
factor  of  95  per  cent,  over  6  miles  of  underground  cable  of  97  per 
cent  conductivity  with  a  loss  of  5  per  cent  of  the  11,000  volts 
generated;  temperature  20°  cent. 

GIVEN  ITEMS. 

W  =  3000  kw.  ;     #0=11  kv.  ;     70  =  5  per  cent; 
L=  6  miles;    K  =  0.95  lag. 

From  Table  31,  page  72,  VJ"  =  4.8. 

From  Table  32,  page  72,  A  =  5L122  =  Q.103. 

0.97 

REQUIRED  ITEMS. 

• 

Size  of  each  wire.  —  From  (66), 

Jf-       (ID*  X  4.8 
0.103  X  3000  X  6 


68  TRANSMISSION  CALCULATIONS 

From  Table  34,  page  73,  use  No.  0000,  for  which  M  =  0.280. 
Per  cent  volt  loss.  —  From  (70), 

V  '"  =  0-280  X  0.103  X  3000  X  6  =  4  30 

(H)2 

From  Table  31,  page  72,  VQ=  4.50  per  cent. 
Volt  loss.  —  From  (73),  v  =  10  X  4.5  X  11  =  495  volts. 
Per  cent  power  loss.  —  From  Table  3,  page  7,  R  =  0.258  ohm. 
From  (75), 

p  _  0.258  X  0.103  X  3000  X  6          ^         cent 
[11  (1  -  0.045)  0.95]2 

Amperes  per  wire.  —  From  Table  33,  page  72,  B  =  0.578. 
From  (79), 

I  =       °'578  ><  3QQQ       =  174  amperes. 
11  (1  -  0.045)  0.95 

From  Table  4,  page  8,  it  is  seen  that  the  cable  will  not  over- 
heat. 

Power-factor  at  source.  —  From   (80), 

K0=  (1  +  0.048)  (1  -  0.045)  0.95=  0.95. 

Example  24.  —  A  three-phase  cable  with  No.  0  conductors  of 
98  per  cent  conductivity  delivers  2500  kw.  at  6000  volts,  95 
per  cent  leading  power-factor  and  60  cycles  per  second,  to  a 
receiver  3  miles  from  the  generator,  the  temperature  of  the  wires 
being  40°  cent. 

GIVEN   ITEMS. 

IF- 1500  kw.;     #=6kv.;      L  =  5  miles;      K=  0.95  lead. 
From  Table  35,  page  74,     M  =  0.475. 

From  Table  32,  page  72,    A  =  -^  =  0.110. 

(J.9o 

REQUIRED  ITEMS. 

Per  cent  volt  loss.  —  From  (69), 

v  _  0.475  X  0.11  X  1500  X  5  _  1Q  9         cent 
(6)2 

Kilovolts  at  source.  —  From  (71), 

#0=  6  (1  +  0.109)  =  6.65  kv. 
Per  cent  power  loss.  —  From  Table  3,  page  7,   R  =  0.518  ohm. 


TRANSMISSION   BY   UNDERGROUND  CABLES  69 


From  (74), 

0.518  X  0.11  X  1500  X  -5 

(6)2  (0.95) 


=    .  .  =  ^         cent 

2 


Amperes  per  wire.  —  From  Table  33,  page  72,  B  =  0.578. 
From  (78), 

7  =  0.578  X  1500 
6  X  0.95 

From  Table  4,  page  8,  this  current  is  slightly  in  excess  of  the 
safe  current-carrying  capacity  of  the  cable. 
Power-factor  at  source.  —  From  (80), 

g       d+0.132)  0.95  a  Q97a 
1  +  0.109 

There  is  no  indication  from  the  above  result  whether  the  cur- 
rent at  the  source  is  leading  or  lagging.  It  then  becomes 
necessary  to  construct  a  vector  diagram  similar  to  Fig.  2,  page  30, 
from  which  it  will  be  found  that  the  current  in  the  above  case  is 
leading  at  the  source. 

Example  25.  —  A  three-phase  load  of  400  kw.  at  95  per  cent 
lagging  power-factor  and  125  cycles  per  second  is  to  be  trans- 
mitted 8  miles  with  a  loss  of  7.5  per  cent  of  the  generated  voltage, 
6600.  Include  capacity  effects  and  assume  wires  of  97  per  cent 
conductivity  and  a  temperature  of  30°  cent. 

GIVEN   ITEMS. 

W=  400  kw.;        EQ=  6.6  kv.;          70=  7.5  per  cent; 
L=8  miles;         K  =0.95  lag. 

From  Table  31,  page  72,   70"'=  6.95. 

From  Table  32,  page  72,       A  -  ^~-  =  0.107 

0.97 

REQUIRED   ITEMS. 

Apparent  power-factor  of  load.  —  Assuming  that  the  required  size 
of  conductors  will  be  between  Nos.  1  and  6,  c=  0.0006  (from 
Table  30,  page  65). 

From  Table  14,  page  39,  the  reactance  factor  for  95  per  cent 
lagging  power-factor  =  0.33. 


70  TRANSMISSION  CALCULATIONS 

From  (44), 

„  0.0006  [6.6  (1  -  0.075)]2  125  X  8          97 

~400~~ 

By  interpolation  in  Table  14,  page  39,  the  apparent  power 
factor  of  load,   K',  =  0.97  lag. 
Size  of  each  wire.  —  From  (66), 


0.107  X  400  X  8 
From  Table  35,  page  74,  use  No.  2  cable,  for  which 

M  =  0.925,  approximately. 
Per  cent  volt  loss.  —  From  (70), 

v  "'  -  0-925  X  0.107  X  400  X  8  _  7  97 
(6.6)2 

From  Table  31,  page  72,  70  =  7.87  per  cent. 

Per  cent  power  loss.  —  From  Table  3,  page  7,   R  =  0.824  ohm. 

From   (75), 


Kilowatts  at  source.  —From  (76),  W0  =  400  (1  +0.081)  =  432  kw. 
Power-factor  at  source  (not  including  capacity).  —  From  (80), 
KQ'=  (1  +  0.081)  (1  -  0.0727)  0.97  =  0.97  lag. 

From  Table  14,  page  39,  t0'  =  0.27. 
From  (45),  page  38, 

n  97       0.0006  (6.6)2  125  X  8  _  ft  91 
t~  ~~ 


From  Table  14,  page  39,  the  'true  power-factor  at  load,   K 
(corresponding  to  £0),  =  0.98  lag. 


TRANSMISSION  BY  UNDERGROUND  CABLES 


71 


Formulas  for  A.  C.  Transmission  by  Underground  Cables. 


Required  Items. 

For  Voltage  Given  at  Load. 

For  Voltage  Given  at  Source. 

Size  of  each  wire 
from 
per  cent  volt  loss 

A  from  Table  32,  page  72. 

&v 

FO"'  from  Table  31,  page  72. 

*-%%L    '•    <*> 

AWL'    '   * 

Then  find  wire  from  Table  34  or  35,  page  73  or  74. 

Size  of  each  wire 
from 
per  cent  power  loss. 

PWK* 

R      P[Ea(l  -  0.01  F0)lTp 

AWL    '•'••' 

AWL 

(68) 

Then  find  wire  from  Table  3,  page  7. 

per  cent  volt  loss. 

v      MA  WL            ,69) 

V»,_XAWL           (70) 

E2 

V0  from  Table  31,  page  72. 

Kilovolts  at  source 
or  load. 

E0  =E(l  +  0.01F).  (71) 

*  =  *0(1-0.0170).(72) 

Volt  loss. 

v  =  1000  (E0  -  E)  =  10  VE  =  10  FO^O  .   .    .  (73) 

per  cent  power  loss. 

R  from  Table  3,  page  7. 

p       RA  WL             ,-£. 

p               RAWL 

(76) 

Kilowatts  at  source. 

WQ=W(l 

+  0.01P)     ....      (76) 

Watt  loss. 

p  =  1000(TF0-TF)  =  10  PW    .   .   .(77) 

Amperes  per  wire. 
(See  par.  34,  page  35.) 

B  from  Table  33,  page  72. 

I-BW                 (78) 

I                BW            (70) 

1      EK     '    '   '    '( 

Power-factor  at  source. 

K  _  (1  +  O.OIP)K  _(l 

f-0.01P)(l-0.01F0)lT.(80) 

1+0.01F 

When  given  E  and  F0,  find  V  from  V  =  V0/  (1  - 
When  given  E0  and  V,  find  V0  from  VQ  =  V  /  (1 
When  given  W',  find  W  from  W  =  W'K     .    .    . 

-  0.01  F0)  .    .    .              (81) 

+  0.01  F)  (82) 
(83) 

NOTATION. 


A  =  Transmission  factor.  Table  32,  page  72.  P  = 

B  =  Transmission  constant.  Table  33,  p.  72.  p  = 

E  =  Kilovolts  (1000  volts)  between  wires  at  R  = 

load.    See  par.  32,  p.  35. 

E0=  Kilovolts  (1000  volts)  between  wires  at  V  = 

source.     See  par.  32,  p.  35.  F0  = 

/  =  Amperes  per  wire.    See  par.  34,  p.  35.  V0'"  = 

K  =  Power-factor  of  load,  in  decimals.  v  = 
K0=  Power-factor  at  source,  in  decimals. 

L  =  Distance  from  source  to  load,  in  miles  W  = 

or  1000  ft.  Wg  = 

M  =  Wire  factor,  Tables  34  and  35,  pp.  73  W*  = 

and  74. 


Power  loss  in  per  cent  of  power  at  load. 

Power  loss  in  watts. 

Resistance    per  mile  of  one  wire,  in 

ohms.    Table  3,  p.  7. 
Volt  loss  in  per  cent  of  volts  at  load. 
Volt  loss  in  per  cent  of  volts  at  source, 
Volt  loss  factor.    Table  31,  p.  72. 
Volt  loss  (volts  at  source  —  volts   a 

load). 

Real  kilowatts  at  load. 
Real  kilowatts  at  source. 
Kilovolt-amperes  (apparent  power)  at 

load. 


72 


TRANSMISSION  CALCULATIONS 


Table  31.— Values  of   F0'"  =  F0  (i-o.oi  F0). 


FO 

FO'" 

Fo 

FO'" 

FO 

FO'" 

FO 

FO'" 

FO 

FO'" 

FO 

FO'" 

V0 

FO'" 

0.0 

0.0 

2.0 

2.0 

4.0 

3.8 

6.0 

5.6 

8.0 

7.4 

10.0 

9.0 

15.0 

12.8 

0.2 
0.4 

0.2 

2.2 

2.1 

4.2 

4.0 

6.2 

5.8 

8.2 

7.5 

10.5 
11.0 

9.4 
9.8 

15.5 
16  0 

13.1 
13.4 

0.6 
0  8 

0.6 
0  8 

2.6 
2  8 

2.5 
2  7 

4.6 
4  8 

4  6 

6  8 

6  3 

8  8 

8  0 

11.5 
12  0 

10.2 
10  6 

16.5 
17  0 

13.8 
14  1 

1.0 

.0 

3.0 

2.9 

5.0 

4.8 

7.0 

6.5 

9.0 

8.2 

12.5 

10.9 

17.5 

14.4 

1.2 

.2 

3.2 

3.1 

5.2 

4.9 

7.2 

6.7 

9.2 

8.4 

13.0 

11.3 

18.0 

14.8 

Table  32.  —  Values  of  A  for  Balanced  Loads. 


2-  Phase  Sys- 

Temperature 
in  Degrees. 

1-Phase  Sys- 
tem with  2 
Equal  Wires. 

2-Phase  Sys- 
tem with  3 
Equal  Wires. 

tem  with  4 
Equal  Wires, 
or  3-  Phase  Sys- 
tem  with   3 

Equal  Wires. 

Cent. 

Fa.hr. 

For  Lin 

Miles. 

For  Lin 
1000ft. 

ForLin 
Miles. 

ForLin 
1000ft. 

ForLin 
Miles. 

ForLin 
1000ft. 

0 

32 

0.185 

0.0351 

0.112 

0.0212 

0.0926 

0.0175 

10 

50 

0.193 

0.0365 

0.116 

0.0220 

0.0963 

0.0182 

20 

68 

0.200 

0.0379 

0.121 

0.0229 

0.100 

0.0190 

30 

86 

0.208 

0.0393 

0.125 

0.0237 

0.104 

0.0197 

40 

104 

0.215 

0.0408 

0.130 

0.0246 

0.108 

0.0204 

50 

122 

0.223 

0.0423 

0.135 

0.0255 

0.112 

0.0212 

The  above  values  are  for  copper  wires  of  100  per  cent  conductivity  and  aluminum  wires  of 
62  per  cent  conductivity,  in  Matthiessen's  standard  scale. 

For  copper  of  other  conductivity,  divide  A  by  given  conductivity. 

For  aluminum  by  other  conductivity,  divide  A  by  given  conductivity  and  multiply 
by  0.62. 

Table  33. —  Values  of  B  for  Balanced  Loads. 


1-Phase  System. 

2-  Phase  System. 

3-  Phase  System. 

1.000 

0.500 

0.578 

TRANSMISSION  BY  UNDERGROUND  CABLES 


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TRANSMISSION  CALCULATIONS 


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CHAPTER   V. 
INTERIOR  WIRES   FOR  ALTERNATING-CURRENT  DISTRIBUTION. 

52.  Introduction.  —  Interior  wiring  involves  short  runs  of  con- 
ductors; therefore  the  conductors  are  often  determined  by  their 
current-carrying  capacity  rather  than  by  conditions  of  maximum 
drop.     It  is  generally  advisable  to  note  the  required  size  of  wire 
for  both  conditions.     Table  4,  page  8,  gives  the  National  Electric 
Code  Standard  for  current-carrying  capacity  of  interior  wires. 

The  formulas  for  interior  wiring  calculations  are  similar  to 
the  preceding  ones  for  alternating-current  transmission.  The 
units  of  power,  voltage  and  distance  have  been  changed  in  order 
to  facilitate  calculation.  All  the  required  items  are  expressed 
in  terms  of  the  current  per  wire  and  the  per  cent  volt  loss  is  also 
given  in  terms  of  the  power  at  load  so  that  problems  involv- 
ing watts  at  load  and  voltage  at  source  may  be  solved  without 
preliminary  approximation. 

The  ordinary  error  of  the  calculation  is  considerably  less  than 
stated  in  paragraph  45  for  underground  conductors,  on  account 
of  the  smaller  wires  usually  employed.  The  error  at  20°  cent, 
of  the  common  assumption  that  the  resistance  drop  equals  the 
volt  loss  is  indicated  in  Table  36,  page  76,  for  wires  in  conduits 
and  on  3-inch  centers. 

It  is  apparent  that  results  based  on  the  assumptions  of 
Table  36  may  be  much  greater  or  much  less  than  the  true  values, 
even  at  power-factors  near  unity.  The  error  is  due  to  the 
power-factor  angles  of  the  load  and  line.  (See  Figs.  1  and  2, 
page  30.) 

53.  Properties  of  Conductors.  —  The  values  of  M  have  been 
calculated  for  wires  of  100  per  cent  Matthiessen's  Standard  at  a 
temperature  of  20°  cent,  or  68°  fahr.     However,  the  effects  of 
temperature  and  conductivity  are  introduced  by  means  of  a  and  b 
in  Table  37,  page  82. 

54.  Spacing  of  Wires.  —  Table  39,  page  83,  gives  the  values  of 
M  for  wires  with  a  thickness  of  insulation  of  ££  inch,  while 

75 


76 


TRANSMISSION  CALCULATIONS 


Table  40,  page  84,  is  calculated  for  conductors  on  3-inch  centers, 
Any  slight  variation  from  either  of  these  spacings  will  not  appre- 
ciably alter  the  results,  even  for  the  largest  wires  at  the  highest 
commercial  frequency. 

Table  39  is  to  be  used  for  wires  in  conduit,  duplex  cables, 
multi-conductor  cables  or  twisted  wires.  Table  40  is  for  wires 
in  molding,  or  for  open  work  such  as  wires  on  cleats  or  knobs. 

55.  Ampere-Feet.  —  A  simple  method  of  dealing  with  dis- 
tributed lamps  is  by  use  of  formula  (84)  on  page  8L  The  term 
II  is  the  ampere-feet  and  is  equal  to  the  sum  of  the  products 
given  by  each  load,  /  multiplied  by  its  distance  I. 

Table  36.  —  Error  in  Per  Cent  of  True  Volt  Loss, 
Assuming  Resistance  Drop  =  Volt  Loss. 


Size    of 
Wire. 
A.W.G. 

60  Cycles  per  Sec. 

125  Cycles  per  Sec. 

Lagging  Power-  Factor  of  Load  in  Per  Cent. 

100 

90 

80 

100 

90 

80 

Wires  in  Conduit. 

0000 
4 
14 

—2 
0 
0 

-15 

+  2 
+  8 

-15 

+  8 
+  18 

-8 
0 
0 

-35 

—  .5 

+  7 

-39 
2 

+  16 

f                               Wires  on  3-Inch  Centers. 

0000 
4 
14 

-7 
—0 
0 

-34 
-4 

+  8 

-37 
-  1 
+  17 

2 
0 

—54 
—16 
+  6 

—59 
—16 

+  14 

56.  Examples.  —  The  application  of  formulas  on  page  81  are 
illustrated  in  the  following  examples.  It  should  be  carefully 
noted  that  the  per  cent  volt  and  power  losses  are  expressed  as 
whole  numbers  and  that  I  is  the  distance  of  transmission  from 
source  to  load,  which  is  the  same  as  the  length  of  one  wire.  In 
formulas  (90)  and  (91),  r  is  the  resistance  per  foot  of  one  wire, 
equal  to  the  values  in  Table  3,  column  4,  divided  by  1000.  The 
meaning  of  the  term  source  is  stated  in  paragraph  32,  page  35. 


WIRES  FOR  ALTERNATING-CURRENT  DISTRIBUTION      77 

Example  26.  —  A  group  of  lamps  having  a  power-factor  of 
95  per  cent  lag  is  to  be  supplied  with  20  amperes  at  60  cycles 
per  second  over  a  single-phase  copper  cable  150  feet  long.  Cal- 
culate the  size  of  wire  for  2  volts  drop  at  30°  cent,  for  copper 
of  97  per  cent  conductivity. 

GIVEN  ITEMS. 

/  =  20  amperes;     v  =  2  volts;    I  =  150  feet;    K  =  0.95  lag. 
From  Table  37,  page  82,  a  =  963  X  0.97  =  934. 

REQUIRED  ITEMS. 

Size  of  each  wire,  —  From  (84), 

934  x  2 


20  X  150 


0.623. 


From  Table  39,  page  83,  use  No.  5,  for  which  M  =  0.623. 

Example  27.  —  At  40°  cent,  a  single-phase  cable  with  No.  8 
conductors  of  97  per  cent  conductivity  and  200  feet  long,  delivers 
25  amperes  at  98  per  cent  lagging  power-factor  from  a  trans- 
former which  gives  100  volts  at  125  cycles  per  second. 

GIVEN  ITEMS. 

/  =  25  amperes;    e0  =  100  volts;    I  =  200  feet;     K  =  0.98  lag. 
From  Table  39,  page  83,  M  =  1.27, 
From  Table  37,  page  82, 

a  =  928  X  0.97  =  900;     b  =  -^  =  222. 

\j  *  y  / 

REQUIRED  ITEMS. 

Volt  loss.  —  From  (85), 

1.27  xsx  200  =  70voltg 


Volts  at  load.  —  From  (89),      e  =  100  -  7  =  93  volts. 
Per  cent  volt  loss.  —  From  (86), 


Per  cent  power  loss.  —  From  Table  3,  page  7,  r  =  0.000627. 


78  TRANSMISSION  CALCULATIONS 

From  (90), 

p  _  0.000627  X  222  X^25  X  200  =  ? 

Watts  at  load.  —  From  Table  38,  page  82,  B  =  1.000. 

From  (94),      w  =  25  X  93  X  °'98  -  2280  watts. 

1  .00 

Watt  loss.  —  From  (92),  p=  0.076  X  2280==  173  watts. 
Power-factor  at  source.  —  From  (96), 


Example  28.  —  A  60-cycle  step-down  transformer  with  120 
volts  at  its  secondary  terminals  is  connected  to  a  load  of  2000 
watts  at  90  per  cent  lagging  power-factor  over  200  feet  of  single- 
phase  circuit  on  cleats.  Determine  the  size  of  wire  for  5  per  cent 
drop  at  20°  cent,  and  copper  of  98  per  cent  conductivity. 

GIVEN  ITEMS. 

w  =  2000  watts;     e0=  120  volts;     70  =  5  per  cent; 
1=  200  feet;        K  =  0.90  lag. 

From  Table  31,  page  72,  70///=  4-8- 

From  (98),       v=  0.05  X  120  =  6  volts. 

From  Table  37,  page  82,        a  =  1000  X  0.98  -  980. 

From  (95),  (in  which  B  from  Table  38,  page  82,  =  1.000), 

2000 


120  (1  -  0.05)  0.90 

REQUIRED  ITEMS. 

Size  of  each  wire.  —  From  (84), 

M=    98QX6    =1.51. 
19.5  X  200 

From  Table  40,  page  84,  use  No.  8,  for  which  M=  1.20. 
Per  cent  volt  loss.  —  From  (87), 

v  '"  =  L2Q  X  200°  X  20°     m  3  40 
0.01  X  980  X  (120)2 


WIRES  FOR  ALTERNATING-CURRENT  DISTRIBUTION      79 

From  Table  31,  page  72,    V0  =  3.50  per  cent. 

Volts  at  load.  — From  (89),  e=  120  (1  -  0.035)  =  116  volts. 

Amperes  per  wire.  —  From  (94), 

=  19.2  amperes. 


116  X  0.90 


From  Table  4,  page  8,  it  is  seen  that  the  current  will  be  safely 
carried  by  No.  8  wires. 

Example  29.  —  A  load  of  1500  watts  at  110  volts,  85  per  cent 
lagging  power-factor  and  60  cycles  per  second  is  to  be  delivered 
250  feet  over  a  two-phase  four- wire  circuit  on  cleats,  with  a  loss 
of  2  per  cent  of  the  generated  volts.  Calculate  the  size  of  wire  of 
98  per  cent  conductivity  for  a  temperature  of  30°  cent. 

GIVEN  ITEMS. 

W  =  1500  watts;     e  =  110  volts;     V0  =  2  per  cent; 
1=  250  feet;        K=0.85  lag. 

From  (98),  v  =  °-02  X  11Q  -  2.25  volts. 

JL   — ~  U.U^  ^ 

From  Table  37,  page  82,  a  =  963  X  0.98  =  944. 

From  (94)  (in  which  B  from  Table  38,  page  82,  =  0.500), 


REQUIRED  ITEMS. 

Size  of  each  wire.  —  From  (84), 

M  =  944  X  2.25 
8  X  250 

From  Table  40,  page  84,  use  No.  8,  for  which  M  =  1.18. 
Per  cent  volt  loss.  —  From  (86), 


Volts  at  source.  —  From  (88),  e0=  110  (1  +  0.0227)  =  113  volts. 


80  TRANSMISSION   CALCULATIONS 

Example  30.  —  A  125  volt,  125  cycle  generator  is  to  supply  a 
three-phase  load  of  100  amperes  per  wire  at  100  per  cent  power- 
factor  over  225  feet  of  cable,  with  a  loss  of  3  per  cent  of  the  gen- 
erated volts.  The  wires  are  to  have  a  conductivity  of  99  per  cent 
and  a  temperature  of  30°  cent. 

GIVEN  ITEMS. 

7  =  100  amperes;     e0=  125  volts;     F0  =  3  per  cent  ; 
/=  225  feet;        K=  LOO. 

From  (98),  v=0.03  X  125=3.75  volts. 
From  Table  37,  page  82, 


a  =  1110  X  0.99=  1100;     &  =       -=181. 

REQUIRED  ITEMS. 

Size  of  each  wire.  —  From  (84), 

,.1100  X  3.75_ft100 
100  X  225  ' 

From  Table  39,  page  77,  use  No.  00,  for  which  M  =  0.163. 
Volt  toss.  —  From  (85), 


Per  cent  volt  loss.  —  From  (98), 


Volts  at  toad  —  From  (89),  e=  125  -  3.3=  121.7  volts. 


WIRES  FOR  ALTERNATING-CURRENT  DISTRIBUTION      81 


Formulas  for  A.C.  Interior  Wiring. 


Required  Items. 

For  Voltage  Given  at  Load. 

For  Voltage  Given  at  Source. 

Size  of  each  wire. 

a  from  Table  37,  page  82. 

*- 

au 

(84) 

II     ' 

Then  find  wire  from  Table  39  or  40,  page  83  or  84. 

Volt  loss. 

v  =  **!! 

a 

(85) 

Per  cent  volt  loss. 

Mil 

y   ,„                       3fW>Z 

0.01  ae02  '  '   * 

(87) 

0.01  ae      O.Ole* 

F0  from  Table  31,  page  72. 

Volts  at  source  or  load. 

eQ  =  e  +  v 
=  e(l  +  0.01F)  .   (88) 

e!e°(l-0.01F0)  . 

(89) 

Per  cent  power  loss. 

r  from  Table  3,  page  7-6  from  Table  37,  page  82. 

*•-!'  <*» 

p                rbll 

(91) 

Watt  loss. 

p  —  0.01  Pw        

(92) 

Watts  at  source. 

wo  =  w  +  p  =  w  (1  +  0.01  P)  . 

(93) 

Amperes  per  wire. 
(See  Par.  34,  page  35.) 

B  from  Table  38,  page  82. 

j       Bw         Bwf         (94) 

I  _            Bw 

(95) 

Bw' 

eK           e 

e0(l-0.01F0)'   ' 

Power-factor  at  source. 

K     (i+o.oip)*:    /nooipvi   OOIFW 

(96) 

1+0.01F 

When  given  F,  find  v  from  v  =  0.01  Fe  =  0.01  Fe0  /(I  +  0.01  F)    .    .   . 
When  given  F0,  find  v  from  v  =  0.01  F()e0  =  0.01  F0e  /  (1  -  0.01  F0)    .    . 
When  given  w  or  w',  find  I  from  (94)  or  (95)  above. 

(97) 
(98) 

NOTATION. 


a  =  Transmission  factor.  Table  37,  page  82.  P  = 

R  =  Transmission  constant.    Table  38,  p.  82.  p  = 

6  =  Transmission  factor.    Table  37,  p.  82.  r  = 
e  =  Volts  between  wires  at  load.    See  par. 

32,  p.  35.  F  = 

c0  =  Volts  between  wires  at  source.  See  par.  F0  = 

32,  p.  35.  FO'"  = 

7  =  Amperes  per  wire.    See  par.  34,  p.  35.  v  = 
K  =  Power-factor  of  load,  in  decimals. 

K0=  Power-factor  at  source,  in  decimals.  w  = 

I  =  Distance  from  source  to  load,  in  feet.  WQ  = 

M=  Wire  factor.     Tables  39  and  40,  pp.  w'  =• 

83  and  84. 


Power  loss  in  percent  of  power  at  load. 

Total  power  loss  in  watts. 

Resistance  per  foot  of  one  wire.  Table 

3,  p.  7. 

Volt  loss  in  per  cent  of  volts  at  load. 
Volt  loss  in  per  cent  of  volts  at  source. 
Volt  loss  factor.    Table  31,  p.  72. 
Volt  loss   (volts   at   source  -  volts  at 

load). 

Real  watts  at  load. 
Real  watts  at  source. 
=  Volt-amperes    (apparent     power)   at 
load. 


82  TRANSMISSION  CALCULATIONS 

Table  37.  —  Values  of  a  and  b  for  Balanced  Loads. 


Temperature 
in   Degrees. 

1  -Phase  Sys- 
tem with  2 
Equal  Wires. 

2-Phase  Sys- 
tem with  3 
Equal  Wires. 

2-Phase  Sys- 
tem with  4 
Equal  Wires. 

3-Phase  Sys- 
tem with  3 
Equal  Wires. 

Cent. 

Fahr. 

a 

6 

a 

b 

a 

6 

a 

b 

0 
10 
20 

32 
50 
68 

1080 
1040 
1000 

185 
193 
200 

895 
860 
828 

185 
193 
200 

1080 
1040 
1000 

185 
193 
200 

1250 
1200 
1160 

160 
167 
173 

30 
40 
50 

86 
104 

T22 

MI 

928 
896 

208 
215 
223 

798 
769 
742 

208 
215 
223 

963 

928 
896 

208 
215 

223 

1110 
1070 
1030 

179 
186 
193 

The  above  values  are  for  copper  of  100  per  cent  conductivity.    For  other  conductivity, 
multiply  a  and  divide  6  by  given  conductivity. 


Table  38.  —  Values  of  B  for  Balanced  Loads. 


1  -Phase  System. 

2-Phase  System. 

3-Phase  System. 

1.000 

0.500 

0.578 

WIRES  FOR  ALTERNATING-CURRENT  DISTRIBUTION      83 


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84 


TRANSMISSION  CALCULATIONS 


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CHAPTER  VI. 
DISTRIBUTION   FOR   SINGLE-PHASE   RAILWAYS. 

57.  Introduction.  —  A  trolley  circuit  consists  primarily  of  an 
overhead  trolley  wire  suspended  from  single,  double  or  multiple 
catenary  construction,  and  the  track  rails.     The  trolley  wire  may 
be  in  parallel  with  an  auxiliary  feeder  or  "  by-pass."     However, 
the  installation  of  auxiliary  feeders  is  the  exception  rather  than 
the  rule,  and  since  their  position  with  respect  to  the  trolley  wire 
and  rails  is  of  much  importance  but  of  wide  variation,  no  attempt 
has  been  made  to  include  them  in  the  results.     The  table  of  M 
has  been  calculated  for  15  and  25-cycle  systems  with  one  and  two 
bonded  rails  per  track.     The  rails  have  a  relative  resistance  of 
steel  to  copper  of  12,  while  the  trolley  wires  are  taken  at  100  per 
cent  conductivity  and  20°  cent. 

58.  Method  of  Calculation.  —  The  value  of  M  is  based  on  the 
"  Report  of  Electric  Railway  Test  Commission  "  and  on  isolated 
calculations   published  in   various   places.     General   laws   have 
been  deduced  for  the  grounded  portion  of  the  circuit,  and  they 
have  been  found  to  agree  sufficiently  well  with  the  experimental 
results.     The  properties  of  the  trolley  and  catenary  construction 
have  been  calculated  and  the  results  have  been  combined  with 
those  for  the  rails  to  obtain  the  values  of  the  complete  circuit. 
The  formulas  on  page  92  are  similar  to  those  for  overhead  trans- 
mission lines  except  that  the  former  have  been  expressed  in  terms 
of  amperes  in  order  to  facilitate  computation.     The  per  cent  volt 
loss  is  also  given  in  terms  of  the  kilowatts  at  load  in  order  to 
avoid   approximation  when   the  voltage   is  known  only  at   the 
source. 

59.  Impedance  of  Rail. —  Due  to  skin  effect,  an  alternating  cur- 
rent flowing  through  a  steel  rail  has  a  diminishing  density  toward 
the  center,  thereby  increasing  the  effective  resistance  of  the  rail. 
Eddy  currents  and  hysteresis  losses  produce  a  further  increase 
in  the  effective  resistance,  resulting  in  an  impedance  to  alter- 
nating current  considerably  greater  than  the  resistance  of  the 

85 


86  TRANSMISSION  CALCULATIONS 

rail  to  direct  current.     The  ratio  of  impedance  divided  by  the 
resistance  to  direct  current  is  called  the  impedance  ratio. 

The  impedance  of  a  complete  track  depends  upon  the  frequency 
of  the  system,  on  the  height  of  the  trolley  wire,  on  the  shape,  size, 
permeability  and  specific  resistance  of  the  rail,  and  on  the  char- 
acter of  bonds  and  roadbed.  It  is  evidently  a  complex  quantity 
and  can  be  determined  only  by  tests  of  installed  track  under 
normal  conditions  of  operation. 

60.  Permeability  of  Rail.  —  The  permeability  of  a  given  rail 
depends  upon  the  density  of  current.     The  method  of  variation 
is  irregular  but  an  average  value  of  the  permeability  can  be 
assumed  without  causing  much  error  in  the  final  result. 

61.  Impedance  and  Weight  of  Rail.  —  The  impedance  depends 
on  the  size  and  shape  of  rail  and  differs  in  the  standard  bullhead 
section  of  foreign  practice  from  the  T-rail  of  this  country.     It 
changes  very  slightly  with  the  weight  of  the  rail  commonly 
installed.     In  any  case  the  impedance  has-been  found  by  experi- 
ment to  vary  inversely  with  the  perimeter  of  the  section. 

62.  Impedance  of  Rail  and  Frequency.  —  Experiment  shows, 
at  least  at  lower  frequencies,  that  the  impedance  of  rails  varies 
directly  as  the  square  root  of  the  cycles  per  second.     Thus  at 
25  cycles  per  second  the  impedance  is  about  1.3  times  the  corre- 
sponding value  at  a  frequency  of  15. 

63.  Formula  for  Rail  Impedance.  —  It  has  been  concluded 
from  experiments  that  the  impedance  of  rails  varies  directly 
as    the   square   root   of    the   frequency   and    inversely   as   the 
perimeter.     However,    the    perimeter    of    standard    T-rails    is 
approximately  proportional  to  the  square  root  of  their  weight 
per  yard.     The   author   has   developed   the   following  formula 
and  has  found  that  it  agrees  sufficiently  well  with  tests  to  serve 
for  purposes  of  practical  calculation. 

Impedance  per  mile  of  one  T-rail,  bonded  and  installed, 


-"v/i 


'Cycles  per  second 
Pounds  per  yard 

The  constant  0.8  is  based  on  an  impedance  ratio  of  6.6  at 
25  cycles  per  second  for  a  75-pound  bonded  rail  of  a  relative 
resistance  of  steel  to  copper  of  about  12,  with  the  trolley  wire 
about  20  feet  above  the  track.* 

*  General  Electric  Co.'s  Bulletin,  No.  4392,  1904. 


DISTRIBUTION  FOR  SINGLE-PHASE  RAILWAYS  87 

64.  Power-Factor  of  Track.  —  The  power-factor  increases  with 
the  current,  but  at  the  lower  frequencies  this  variation  is  small 
for  normal   current   densities.     Tests   of   the   Electric   Railway 
Commission   show   a   slight  increase   of  power-factor  with  the 
weight  of  rail  but  other  experiments  indicate  practically  no 
change.*     The   power-factor  of   the   track   has   been    assumed 
constant  for  all  rail  weights,  at  65  per  cent  for  15  cycles  and  55 
per  cent  for  25  cycles  per  second. 

65.  Height     of    Trolley.  —  Trolley     wires     for     single-phase 
systems  range  in  height  from  about  18  to  22  feet  above  rails. 
The  height  assumed  in  the  calculations  of  M  is  22  feet,  but  any 
variation  within  standard  limits  will  have  small  effect  on  the 
impedance  of  the  trolley,  and  almost  no  effect  on  the  impedance 
of  the  track. 

66.  Effect   of  Catenary   Construction.  —  The   trolley   wire  is 
supported  from  a  single  catenary  wire,  or  from  two  wires  arranged 
either  horizontally  or  vertically.     In  almost  every  case  the  trolley 
and  catenary  wires  are  electrically  connected.     However,  since 
the  catenary  wires  are  steel,  their  skin  effect  prevents  them  from 
carrying  more  than  a  small  part  of  the  current,  which  for  single 
catenary  construction  has  been  determined  by  calculation  to 
be  equivalent  to  about  10  per  cent.f     In  double  or  multiple 
catenary  construction  the  proportional  part  of  the  current  carried 
by  the  trolley  is  somewhat  reduced,  but  in  order  to  be  safe  in 
calculation  for  either  type  of  overhead  construction,  the  above 
result  has  been  used. 

67.  Impedance  of  Complete  Circuit.  —  The  ohmic  resistance 
and  the  reactance  of  the  track  and  trolley  having  been  calcu- 
lated, the  two  were  combined  to  obtain  the  total  impedance. 
Table  42,  page  93,  gives  the  values  of  M.     It  will  be  observed 
that  the  effect  of  varying  the  rail  weight  is  less  pronounced  than 
of  varying  the  size  of  trolley,  or,  in  other  words,  that  the  greater 
part  of  the  total  impedance  is  due  to  the  trolley.     Since  the 
impedance  of  the  overhead  system  is  susceptible  of  fairly  accurate 
calculation,  it  follows  that  any  error  and  variation  in  the  assumed 
value  of  the  track  impedance  will  have  a  small  effect  on  the  final 
result.     This  conclusion  is  corroborated  by  tests. 

Following  is  a  comparison  between  the  calculated  impedance 

*  Foster's  "Electrical  Engineer's  Pocketbook,"  1908,  page  795. 
t  J,  B.  Whitehead,  Proceedings  of  A.I.E.E.,  May,  1908,  page  627. 


88  TRANSMISSION  CALCULATIONS 

per  mile  and  the  published  results  of  tests  on  installed  lines. 
The  values  of  the  Electric  Railway  Test  Commission  are  not 
included,  as  their  test  track  is  not  similar  to  modern  construction. 

Table  41.  —  Test  and  Calculated  Values  of  Impedance 
Per  Mile  at  25  Cycles  Per  Second. 


Rails. 

Trolleys. 

Test 
Values. 

Calculated 
Values. 

4-75  Ib. 
2-70  Ib. 
2-80  Ib. 

2  No.  000 
1  No.  000 

1  No.  000 

0.3!  (a) 
0.60  (b) 
0.65  to  0.70  (c) 

0.34 
0.68 
0.67 

(a)  General  Electric  Co.'s  Tests  on  Ballston  Line,  Bulletin  4392,  1904. 

(b)  Based  on  Westinghouse  Tests,  Foster's  "Electrical  Engineer's  Pocketbook,"   1908 
page  797. 

(c)  J.  B.  Whitehead,   Proceedings  of  A.  I.  E.  E.,  May,   1908,   page  638;  based  on 
tests  and  calculations. 

68.  Multiple  Tracks.  —  With  small  error  the  results  may  be 
applied  to  a  number  of  tracks,  either  by  proportioning  the  total 
load  or  by  dividing  the  wire  factor  in  Table  42  for  one  track 
by  the  number  of  tracks.  Thus  the  value  of  M  at  25  cycles  per 
second  and  80  per  cent  load  power-factor  for  four  80-pound 
rails  and  two  No.  000  trolleys  is 

=.0.0325. 


69.  Examples.  - —  For  convenience  in  computation  the  for- 
mulas have  been  expressed  in  terms  of  current.  The  per  cent 
volt  loss  has  also  been  given  in  terms  of  kilowatts  at  the  load, 
so  that  no  preliminary  approximation  is  required  when  the  volt- 
age is  known  only  at  the  source.  The  meaning  of  the  term 
"  source  "  is  stated  in  paragraph  32,  page  35. 

The  application  of  the  formulas  on  page  92  is  illustrated  in 
the  following  typical  examples. 

Example  31.  —  Calculate  the  volt  loss  per  mile  per  100  amperes 
at  a  load  power-factor  of  80  per  cent  for  15  and  25-cycle  single- 
phase  railway  circuits  as  follows: 

Two  70-lb.  rails  and  one  No.  00  trolley. 
Four  80-lb.  rails  and  two  No.  000  trolleys. 
Eight  100-lb.  rails  and  four  No.  0000  trolleys. 


DISTRIBUTION  FOR  SINGLE-PHASE  RAILWAYS 


89 


From  (107)  and  Table  42,  page  93,  the  following  results  are 
obtained: 


Volt  Loss  per  Mile  per  100  Amperes. 

Rails. 

Trolleys. 

15  Cycles  per  Sec. 

25  Cycles  per  Sec. 

2-70  lb. 

1  No.  00 

61.0 

73.0 

4-80  lb. 

2  No.  000 

26.0 

32.5 

8-100  lb. 

4  No.  0000 

11.5 

14.5 

Example  32.  —  A  25-cycle  single-phase  railway  system  having 
two  70-lb.  bonded  rails  is  to  be  designed  for  a  loss  of  10  per 
cent  of  the  3300  generated  volts  when  delivering  150  amperes  at 
90  per  cent  power-factor  to  a  car  4  miles  away. 

GIVEN   ITEMS. 

7  =  150  amperes;     EQ=  3.3  kv.;       V0  =  10  per  cent; 
L  =  4  miles;       K  =  0.90  lag. 

REQUIRED  ITEMS. 

Size  of  trolley  and  rail.  —  From  (101), 


M 


3.3  X  10 


0.055. 


150  X  4 

From  Table  42,  page  93,  use  No.  0000  trolley  wire,  for  which 
=  0.057. 

Per  cent  volt  loss.  —  From  (103), 

Tr       0.057  X  150  X  4 


3.3 


=  10.4  per  cent. 


Kilovolts  at  car.  —  From  (105),  E  =  3.3  (1  -  0.104)  =  2.96  kv. 
Per  cent  power  loss. — From  Table  42,  page  93,  72  =  0.364  ohm. 
From  (109), 

0.364  X  150  X  4 


P  = 


10  X  3.3  (1  -  0.104)  0.90 


8.2  per  cent. 


Power-factor  at  source.  —  From  (114), 

K0  =  (l'+  0.082)  (1  -  0.104)  0.90=  0.87  lag. 


90  TRANSMISSION  CALCULATIONS 

Example  33.  —  A  25-cycle  single-phase  road  with  four  80- 
Ib.  bonded  rails  and  two  No.  000  trolleys  supplies  each  of 
three  cars  with  50  amperes  at  85  per  cent  power-factor  when 
located  at  3,  5,  and  10  miles  away,  respectively,  the  power  station 
voltage  being  11,000. 

Owing  to  the  power-factor  at  the  first  two  cars,  the  following 
solution  is  slightly  in  error. 

GIVEN    ITEMS. 

IL  =  50  (3  +  5  +  10)  =  900  ampere-miles. 


From  Table  42,  page  93,    M  =  =  0.0315. 

REQUIRED    ITEMS. 

Per  cent  volt  loss  to  last  car.  —  From  (103), 
FQ=  0.0315  X  900  =26pe 

Kilovolts  at  last  car.  —  From  (106),  #=11  (1-0.026)  =10.7  kv. 

Example  34.  —  A  25-cycle  single-phase  car  starting  8  miles 
from  a  transformer  station  generating  6600  volts  takes  500  kw. 
at  80  per  cent  power-factor  from  a  circuit  consisting  of  two  70- 
Ib.  rails  and  one  No.  000  trolley. 

GIVEN  ITEMS. 

TF=500kw.;     #0=6.6kv.;     L=  8  miles;      K=  0.80  lag. 
From  Table  42,  page  93,     M  =  0.066. 

REQUIRED  ITEMS. 

Per  cent  volt  loss.  —  From  (104), 

v  „,  =  0.066  X  500  X  8  _  -  fi 
(6.6)2  0.80 

From  Table  31,  page  72,  70=  8.3  per  cent. 

Kilovolts  at  load.  —  From  (106),  E  =  6.6  (1  -  0.083)  =  6.05  kv. 

Amperes.  —  From  (112), 

103  amperes 


6.05  X  0.80 
Per  cent  power  loss.  —  From  Table  42,  page  93,  R  =  0.425  ohm. 


DISTRIBUTION  FOR  SINGLE-PHASE  RAILWAYS  91 


Kilowatts  at  source.  —  From  (111), 

W0=  500  (1  +  0.072)  =  536  kw. 

Power-factor  at  source.  —  From  (114), 

K0=  (1  +  0.072)  (1  -  0.083)  0.80  =  0.79  lag. 

Example  35.  —  Two  15-cycle  single-phase  locomotives  start 
simultaneously  10  miles  from  a  power  station  generating  11,000 
volts.  Determine  the  line  voltage  at  the  locomotives  if  each 
takes  3000  kv.-amp.  (kilo  volt-amperes)  at  75  per  cent  power- 
factor,  over  a  circuit  consisting  of  eight  100-lb.  rails  and  four 
No.  0000  trolley. 

GIVEN  ITEMS. 

IF'  =  6000  kv.-amp.;     #0=llkv.;     L=  10  miles;    K  =  0.75  lag. 
From  Table  42,  page  93, 

M  =  ™46  =  0.0115. 
4 

W  -  6000  X  0.75  =  4500  kw. 

REQUIRED  ITEMS. 

Per  cent  volt  loss.  —  From  (104), 

T/  „/  _  0.0115  X  4500  X  10  _  c  7 
0.75 


From  Table  31,  page  72,  70  =  6.1  per  cent. 
Kilovolts  at  locomotives.  —  From  (106), 

tf=ll(l  -  0.061)=  10.3  kv. 


92  TRANSMISSION  CALCULATIONS 

Formulas  for  Single-Phase  Railway  Circuits. 


Required  Items. 

For  Voltage  Given  at  Load. 

For  Voltage  Given  at  Source. 

Size  of  trolley  and  rail. 

M  =  (100) 

jf-SZ..      .(ion 

Per  cent  volt  loss. 

V      MIL                     c\(\f>\ 

v      MIL 

E    ( 

°         Eo       '     '    '  (' 

y    ,,,               MWL 

0           E*K  ' 

V(}  from  Table  31,  page  72. 

Kilovolts  at  source  or 
load. 

E0  =  E(l  +  0.01  F)  .  (105) 

E=E0(1  -001F0)  .  (106) 

Volt  loss. 

v  =  1000(#0  -  E)  =  IOVE  =  WV0E0  =  10  MIL.  (107) 

Per  cent  power  loss. 

R  from  Table  42,  page  93. 

p         RIL                  (-\OR\ 

P                 RIL 

10J£0(1  -  0.01F0).K" 
(109 

Watt  loss. 

p  =  WPW=RPL       (110) 

Kilowatts  at  source. 

WQ=  W  +  0.001  p  =  W(l  +  0.01  P)    (Ill) 

Amperes. 

'-£-¥  •  -(m) 

I         w 

~w> 

(113) 

Power-factor  at  source. 

TT       (1+001P)JK"     ,j  1  A  oi  P)  n      0  Q1FN7T   (114) 

1  -f  0.01  F 

When  given  E  and  F0,  find  F  from  F  =  F0/  (1  -  .01F0)     (115) 
When  given  E0  and  F,  find  F0  f  rorn  F0  =  F/(l  +  .OIF)     (116) 
When  given  W  or  W,  find  I  from  (112)  or  (113)  above. 

NOTATION. 

E      =  Kilovolts  between  trolley  and  rail,  at  load. 
E0     =  Kilovolts  between  trolley  and  rail,  at  source. 
/       =  Total  amperes  from  source. 
K      =  Power-factor  of  load,  in  decimals. 
/To     =  Power-factor  at  source,  in  decimals. 
L       —  Distance  from  source  to  load,  in  miles. 
M      =  Wire  factor.     Table  42,  page  93. 
p      =  Power  loss  in  per  cent  of  power  at  load. 
Power  loss  in  watts. 

Resistance  per  mile  of  circuit.    Table  42,  page  93. 
Volt  loss  in  per  cent  of  volts  at  load. 

V0     =  Volt  loss  in  per  cent  of  volts  at  source. 
/Y"    =  Volt  loss  factor.    Table  31,  page  72. 
v        =  Total  loss  in  volts  (volts  at  source  -  volts  at  load). 

W     =  Real  kilowatts  at  load. 

Wn    =  Real  kilowatts  at  source. 

W    =  Kilovolt-amperes  (apparent  power)  at  load. 


I  : 

v    = 


DISTRIBUTION  FOR  SINGLE-PHASE  RAILWAYS 


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